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4. The quiz participants were asked four questions: 90 participants answered the first question correctly, 50 answered the second, 40 answered the third, and 20 answered the fourth, and no one was able to answer more than two questions correctly. What is the minimum number of participants in the quiz under these conditions? | The total number of answers is 200. Since no one answered more than two questions, the minimum possible number of participants in the quiz is 100, and in this case, each participant in the quiz must answer exactly 2 questions correctly. We will provide an example of a quiz where the described situation is realized. Number the participants from 1 to 100. Let participants numbered from 1 to 90 answer the first question, from 91 to 100 and from 1 to 40 answer the second question, from 41 to 80 answer the third question, and from 81 to 100 answer the fourth question. | 100 | Combinatorics | olympiads | null | null | \(\boxed{100}\) | 4. The quiz participants were asked four questions: 90 participants answered the first question correctly, 50 answered the second, 40 answered the third, and 20 answered the fourth, and no one was able to answer more than two questions correctly. What is the minimum number of participants in the quiz under these conditions?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The total number of answers is 200. Since no one answered more than two questions, the minimum possible number of participants in the quiz is 100, and in this case, each participant in the quiz must answer exactly 2 questions correctly. We will provide an example of a quiz where the |
. On the faces of a six-faced dice, the numbers 1, 2, 3, 4, 5, and 6 are placed. The dice is thrown and falls on the table. After this, the numbers on all faces except one are visible. The numbers on the five visible faces are multiplied. Find the probability that this product is divisible by 16.
Answer: $0.5$. | If an odd number is on the invisible face, then in the product of the remaining five numbers, 2, 4, and 6 are present, and it is divisible by 16. If an even number is on the invisible face, the product of the remaining five digits is not divisible by 16 - its prime factorization will not have four twos. | 0.5 | Number Theory | olympiads | null | null | \(\boxed{0.5}\) | . On the faces of a six-faced dice, the numbers 1, 2, 3, 4, 5, and 6 are placed. The dice is thrown and falls on the table. After this, the numbers on all faces except one are visible. The numbers on the five visible faces are multiplied. Find the probability that this product is divisible by 16.
Answer: $0.5$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If an odd number is on the invisible face, then in the product of the remaining five numbers, 2, 4, and 6 are present, and it is divisible by 16. If an |
. Find the number of natural numbers not exceeding 2022 and not belonging to either the arithmetic progression $1,3,5, \ldots$ or the arithmetic progression $1,4,7, \ldots$ | These two progressions define numbers of the form $1+2n$ and $1+3n$. This indicates that the desired numbers are of the form $6n$ and $6n-4, n \in \mathbb{N}$. Since 2022 is divisible by 6, the numbers of the form $6n$ will be $\frac{2022}{6}=337$, and there will be as many numbers of the form $6n-4$. Therefore, the total number of numbers is $337 \cdot 2=674$. | 674 | Number Theory | olympiads | null | null | \(\boxed{674}\) | . Find the number of natural numbers not exceeding 2022 and not belonging to either the arithmetic progression $1,3,5, \ldots$ or the arithmetic progression $1,4,7, \ldots$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
These two progressions define numbers of the form $1+2n$ and $1+3n$. This indicates that the desired numbers are of the form $6n$ and $6n-4, n \in \mathbb{N}$. Since 2022 is divisibl |
. Find the three last digits of the number $10^{2022}-9^{2022}$. | Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10-1$, then $A(\bmod 1000) \equiv-C_{2022}^{2} \cdot 100+C_{2022}^{1} \cdot 10-1(\bmod 1000) \equiv$ $-\frac{2022 \cdot 2021 \cdot 100}{2}+20220-1(\bmod 1000) \equiv-100+220-1 \equiv 119$. | 119 | Number Theory | olympiads | null | null | \(\boxed{119}\) | . Find the three last digits of the number $10^{2022}-9^{2022}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10 |
5.1. (12 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, acting solely in his own interest, would definitely pay the guard? | If the guard asks for 199 coins, then the outsider, agreeing, will give him this amount, but will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If the guard demands 200, the outsider might refuse, as there is no difference in profit. If the guard demands more, it is more profitable for the outsider to refuse. The guard can ask for less, but the problem requires finding the largest amount. Thus, the answer is 199 coins. | 199 | Logic and Puzzles | olympiads | null | null | \(\boxed{199}\) | 5.1. (12 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offered him to name a sum. What is the largest number of coins the guard can ask for so that the stranger, acting solely in his own interest, would definitely pay the guard?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the guard asks for 199 coins, then the outsider, agreeing, will give him this amount, but will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one |
10.1. (12 points) The surface of a round table is divided into $n$ identical sectors, in which numbers from 1 to $n (n \geqslant 4)$ are written sequentially clockwise. Around the table sit $n$ players with numbers $1,2, \ldots, n$, going clockwise. The table can rotate around its axis in both directions, while the players remain in place. The players are seated at equal distances from each other, so when the table stops rotating, exactly one player is opposite each sector and receives the number of coins written on that sector. After $m$ rotations of the table, player №1 received 74 fewer coins than player №4, and player №2 received 50 fewer coins than player №3. Find $m$, given that player №4 received 3 coins twice as often as 2 coins, but half as often as 1 coin. | Let player No. 3 get one coin exactly $k$ times, then we get the equation $(m-k)-(n-1) k=50$. Exactly in $7 k$ cases, one coin was received by someone from players $2,3,4$, then we get the equation $3(m-7 k)-7 k(n-3)=74$. We expand the brackets, combine like terms, and get a system of two equations $m-n k=50,3 m-7 n k=74$. We multiply the first equation by 7, subtract the second from it, and get $m=69$. | 69 | Combinatorics | olympiads | null | null | \(\boxed{69}\) | 10.1. (12 points) The surface of a round table is divided into $n$ identical sectors, in which numbers from 1 to $n (n \geqslant 4)$ are written sequentially clockwise. Around the table sit $n$ players with numbers $1,2, \ldots, n$, going clockwise. The table can rotate around its axis in both directions, while the players remain in place. The players are seated at equal distances from each other, so when the table stops rotating, exactly one player is opposite each sector and receives the number of coins written on that sector. After $m$ rotations of the table, player №1 received 74 fewer coins than player №4, and player №2 received 50 fewer coins than player №3. Find $m$, given that player №4 received 3 coins twice as often as 2 coins, but half as often as 1 coin.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let player No. 3 get one coin exactly $k$ times, then we get the equation $(m-k)-(n-1) k=50$. Exactly in $7 k$ cases, one coin was received by someone from players $2,3,4$, then we get the equation $3(m- |
3. Find the minimum value of the expression $\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}}$ under the conditions $|x|<1,|y|<2$ and $x y=1$.
Answer: 4. | Using the inequality between the arithmetic mean and the geometric mean, under the given conditions on $x$ and $y$, we obtain
$$
\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}} \geqslant 2 \sqrt{\frac{1}{1-x^{2}} \cdot \frac{4}{4-y^{2}}}=\frac{4}{\sqrt{5-4 x^{2}-y^{2}}}=\frac{4}{\sqrt{1-(2 x-y)^{2}}} \geqslant 4
$$
and all inequalities become equalities if $2 x=y$ and $x y=1$, i.e., $x=\frac{1}{\sqrt{2}}, y=\sqrt{2}$. | 4 | Algebra | olympiads | null | null | \(\boxed{4}\) | 3. Find the minimum value of the expression $\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}}$ under the conditions $|x|<1,|y|<2$ and $x y=1$.
Answer: 4.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Using the inequality between the arithmetic mean and the geometric mean, under the given conditions on $x$ and $y$, we obtain
$$
\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}} \geqslant 2 \sqrt{\frac{1}{1-x^{2}} \cdo |
4. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the ratio of the car's speed to the pedestrian's speed. | In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he would have $1 / 5$ of the bridge's length left to walk, while according to the problem, the car would have reached the beginning of the bridge in time $t$ and would have the entire bridge left to travel before meeting the pedestrian. Thus, the ratio of the car's speed to the pedestrian's speed is 5. | 5 | Algebra | olympiads | null | null | \(\boxed{5}\) | 4. Having walked $2 / 5$ of the length of a narrow bridge, a pedestrian noticed that a car was approaching the bridge from behind. Then he walked back and met the car at the beginning of the bridge. If the pedestrian had continued walking forward, the car would have caught up with him at the end of the bridge. Find the ratio of the car's speed to the pedestrian's speed.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In the time $t$ that the pedestrian walked towards the car until they met at the beginning of the bridge, he covered $2 / 5$ of the bridge's length. Therefore, if the pedestrian continued walking forward, in time $t$ he would have covered another $2 / 5$ of the bridge's length, and he would h |
9. To what power must the root $x_{0}$ of the equation $x^{11} + x^{7} + x^{3} = 1$ be raised to obtain the number $x_{0}^{4} + x_{0}^{3} - 1 ?$ | If $x_{0}=1$, then $x_{0}^{4}+x_{0}^{3}-1=1$, so in this case the degree can be any. But the number $x_{0}=1$ does not satisfy the equation $x^{11}+x^{7}+x^{3}=1$, therefore $x_{0} \neq 1$.
Since $1=x_{0}^{11}+x_{0}^{7}+x_{0}^{3}$, we get
$$
x_{0}^{4}+x_{0}^{3}-1=x_{0}^{4}+x_{0}^{3}-x_{0}^{11}-x_{0}^{7}-x_{0}^{3}=x_{0}^{4}\left(1-x_{0}^{7}-x_{0}^{3}\right)=x_{0}^{4}\left(x_{0}^{11}+x_{0}^{7}+x_{0}^{3}-x_{0}^{7}-x_{0}^{3}\right)=x_{0}^{15} .
$$ | 15 | Algebra | olympiads | null | null | \(\boxed{15}\) | 9. To what power must the root $x_{0}$ of the equation $x^{11} + x^{7} + x^{3} = 1$ be raised to obtain the number $x_{0}^{4} + x_{0}^{3} - 1 ?$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If $x_{0}=1$, then $x_{0}^{4}+x_{0}^{3}-1=1$, so in this case the degree can be any. But the number $x_{0}=1$ does not satisfy the equation $x^{11}+x^{7}+x^{3}=1$, therefore $x_{0} \neq 1$.
Since $1=x_{0}^{11}+x_{0}^{7}+x_{ |
8. 99 wise men sat at a round table. They know that fifty of them are wearing hats of one of two colors, and the other forty-nine are wearing hats of the other color (but it is not known in advance which of the two colors 50 hats are, and which 49 are). Each wise man can see the colors of all the hats except his own. All the wise men must simultaneously write down (each on their own piece of paper) the color of their hat. Can the wise men agree in advance to answer in such a way that at least 74 of them give the correct answer? (U. Feige, proposed by K. Knop) | Let there be 50 white and 49 black hats among the hats. It is clear that the 49 sages who see 50 white and 48 black hats know that they are wearing black hats. Now let each of those who see 49 white and black hats name the color that predominates among the 49 people following them clockwise. If A is one of these sages, and B is the 25th sage from A clockwise wearing a white hat. If there are no more than 48 people between A and B, A will say that he is wearing a white hat; otherwise, B will say that he is wearing a white hat. Since all 50 sages in white hats are divided into 25 such pairs (A, B), 25 of them will correctly name the color of their hat, and in the end, there will be at least $49+25=74$ correct answers. | 74 | Logic and Puzzles | olympiads | null | null | \(\boxed{74}\) | 8. 99 wise men sat at a round table. They know that fifty of them are wearing hats of one of two colors, and the other forty-nine are wearing hats of the other color (but it is not known in advance which of the two colors 50 hats are, and which 49 are). Each wise man can see the colors of all the hats except his own. All the wise men must simultaneously write down (each on their own piece of paper) the color of their hat. Can the wise men agree in advance to answer in such a way that at least 74 of them give the correct answer? (U. Feige, proposed by K. Knop)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let there be 50 white and 49 black hats among the hats. It is clear that the 49 sages who see 50 white and 48 black hats know that they are wearing black hats. Now let each of those who see 49 white and black hats name the color that predominates among the 49 people following them clockwise. If A is one of these sages, and B is the 25th sage from A clockwise w |
. (10 points) A numerical sequence is given:
$x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1$.
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2021$. | # Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check this:
$$
\begin{aligned}
& \left(x_{0}+x_{1}+\ldots+x_{k}\right)+x_{k+1}=\frac{1}{n-k}+x_{k+1}=\frac{1}{n-k}+\frac{1}{n-(k+1)}\left(x_{0}+x_{1}+\ldots+x_{k}\right)= \\
& =\frac{1}{n-k}+\frac{1}{(n-k-1)(n-k)}=\frac{1}{n-(k+1)} .
\end{aligned}
$$
Thus, this equality also holds for $k+1$. Therefore, our assumption that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$ is correct.
Then, for $n=2021$ and $k=n-1=2020: S_{2021}=x_{0}+x_{1}+\ldots+x_{2020}=\frac{1}{2021-2020}=1$. | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | . (10 points) A numerical sequence is given:
$x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1$.
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2021$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check this:
$$
\begin{aligned}
& \left(x_{0}+x_{1}+\ldots+x_{k}\right)+x_{k+1}=\frac |
. (10 points) A numerical sequence is given:
$$
x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1
$$
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2022$. | # Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check this:
$$
\begin{aligned}
& \left(x_{0}+x_{1}+\ldots+x_{k}\right)+x_{k+1}=\frac{1}{n-k}+x_{k+1}=\frac{1}{n-k}+\frac{1}{n-(k+1)}\left(x_{0}+x_{1}+\ldots+x_{k}\right)= \\
& =\frac{1}{n-k}+\frac{1}{(n-k-1)(n-k)}=\frac{1}{n-(k+1)} .
\end{aligned}
$$
Thus, this equality also holds for $k+1$. Therefore, our assumption that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$ is correct.
Then, for $n=2022$ and $k=n-1=2021: S_{2022}=x_{0}+x_{1}+\ldots+x_{2021}=\frac{1}{2022-2021}=1$. | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | . (10 points) A numerical sequence is given:
$$
x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1
$$
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2022$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check this:
$$
\begin{aligned}
& \left(x_{0}+x_{1}+\ldots+x_{k}\right)+x_{k+1}=\frac |
. By how many units can the city's fleet of natural gas vehicles be increased in 2022, assuming that the capacity of each of the old CNG stations in the city is equal to the capacity of the new station on Narodnaya Street, and that the city's fleet constitutes only $70 \%$ of all vehicles refueling at CNG stations in St. Petersburg. Assume that the stations are used at $100 \%$ of their capacity. | Task 2. There are a total of 15 stations: 4 new ones and 11 old ones.
The throughput capacity of the old stations is 11 x $200=2200$ vehicles per day, and for the new ones: $200+700=900$ vehicles per day. In total, the stations can refuel: $2200+900=3100$ vehicles per day.
Vehicles from the city fleet account for only $70 \%$ of all vehicles refueled at the CNG stations in St. Petersburg, so $3100 x 0.7=2170$ vehicles can be refueled. Currently, there are 1000 vehicles in the city fleet, so the city fleet of gas-powered vehicles can be increased by 1170 units. | 1170 | Algebra | olympiads | null | null | \(\boxed{1170}\) | . By how many units can the city's fleet of natural gas vehicles be increased in 2022, assuming that the capacity of each of the old CNG stations in the city is equal to the capacity of the new station on Narodnaya Street, and that the city's fleet constitutes only $70 \%$ of all vehicles refueling at CNG stations in St. Petersburg. Assume that the stations are used at $100 \%$ of their capacity.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Task 2. There are a total of 15 stations: 4 new ones and 11 old ones.
The throughput capacity of the old stations is 11 x $200=2200$ vehicles per day, and for the new ones: $200+700=900$ vehicles per day. In total, the stations can refuel: $2200+900=3100$ vehicles per day.
Vehicles |
6. In a perfectly competitive market, the demand function for a certain good is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=150-\mathrm{p}$, and the supply function for this good is: $\mathrm{Q}_{\mathrm{s}}(\mathrm{p})=3 \mathrm{p}-10$. As a result of a sharp increase in the number of consumers of this good, under all other equal conditions, the demand for it increased by a factor of $\alpha$ at every possible price level. As a result, the price of the good increased by $25 \%$. Find the value of $\alpha$. (8 points) | # Solution:
The new demand function will be $Q_{d}^{\text {new }}(p)=a(150-p)$.
Find the new equilibrium price from the condition of equality of demand and supply: $3 \mathrm{p}-10=\mathrm{Q}_{s}(\mathrm{p})=\mathrm{Q}_{d}^{\text {new }}(\mathrm{p})=a(150-\mathrm{p}): \quad \mathrm{p}^{\text {new }}=\frac{150 a+10}{3+a}$. Note that the initial price $\mathrm{p}_{0}=40$ (when $\alpha=1$). Then, knowing that the price increased by $25 \%$, we write the equation $\mathrm{p}^{\text {new }}=\frac{150 \alpha+10}{3+\alpha}=40 \cdot(1+0.25)=50$. We get, $a=1.4$.
## Grading Criteria:
Correctly found the old equilibrium price - 2 points.
Correctly formulated the equation to find $\alpha$ - 4 points.
Correctly found the value of $\alpha$ - 2 points. | 1.4 | Algebra | olympiads | null | null | \(\boxed{1.4}\) | 6. In a perfectly competitive market, the demand function for a certain good is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=150-\mathrm{p}$, and the supply function for this good is: $\mathrm{Q}_{\mathrm{s}}(\mathrm{p})=3 \mathrm{p}-10$. As a result of a sharp increase in the number of consumers of this good, under all other equal conditions, the demand for it increased by a factor of $\alpha$ at every possible price level. As a result, the price of the good increased by $25 \%$. Find the value of $\alpha$. (8 points)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution:
The new demand function will be $Q_{d}^{\text {new }}(p)=a(150-p)$.
Find the new equilibrium price from the condition of equality of demand and supply: $3 \mathrm{p}-10=\mathrm{Q}_{s}(\mathrm{p})=\mathrm{Q}_{d}^{\text {new }}(\mathrm{p})=a(150-\mathrm{p}): \quad \mathrm{p}^{\text {new }}=\frac{150 a+10}{3+a}$. Note that the initial price $\mathrm{p}_{0}=40$ (wh |
6. In a perfectly competitive market, the demand function for a certain good is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=150-\mathrm{p}$, and the supply function for this good is: $\mathrm{Q}_{\mathrm{s}}(\mathrm{p})=3 \mathrm{p}-10$. As a result of a sharp increase in the number of consumers of this good, the demand for it increased by a factor of $\alpha$ at every possible price level, under all other equal conditions. As a result, the price of the good increased by $25 \%$. Find the value of $\alpha$. (8 points) | # Solution:
The new demand function will be $Q_{d}^{\text {new }}(p)=a(150-p)$.
Find the new equilibrium price from the condition of equality of demand and supply: $3 \mathrm{p}-10=\mathrm{Q}_{s}(\mathrm{p})=\mathrm{Q}_{d}^{\text {new }}(\mathrm{p})=a(150-\mathrm{p}): \quad \mathrm{p}^{\text {new }}=\frac{150 a+10}{3+a}$. Note that the initial price $\mathrm{p}_{0}=40$ (when $\alpha=1$). Then, knowing that the price increased by $25 \%$, we write the equation $\mathrm{p}^{\text {new }}=\frac{150 \alpha+10}{3+\alpha}=40 \cdot(1+0.25)=50$. We get, $a=1.4$.
## Grading Criteria:
Correctly found the old equilibrium price - 2 points.
Correctly formulated the equation to find $\alpha$ - 4 points.
Correctly found the value of $\alpha$ - 2 points. | 1.4 | Algebra | olympiads | null | null | \(\boxed{1.4}\) | 6. In a perfectly competitive market, the demand function for a certain good is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=150-\mathrm{p}$, and the supply function for this good is: $\mathrm{Q}_{\mathrm{s}}(\mathrm{p})=3 \mathrm{p}-10$. As a result of a sharp increase in the number of consumers of this good, the demand for it increased by a factor of $\alpha$ at every possible price level, under all other equal conditions. As a result, the price of the good increased by $25 \%$. Find the value of $\alpha$. (8 points)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution:
The new demand function will be $Q_{d}^{\text {new }}(p)=a(150-p)$.
Find the new equilibrium price from the condition of equality of demand and supply: $3 \mathrm{p}-10=\mathrm{Q}_{s}(\mathrm{p})=\mathrm{Q}_{d}^{\text {new }}(\mathrm{p})=a(150-\mathrm{p}): \quad \mathrm{p}^{\text {new }}=\frac{150 a+10}{3+a}$. Note that the initial price $\mathrm{p}_{0}=40$ (wh |
# Problem 1. Maximum 15 points
A company that produces educational materials for exam preparation incurs average costs per textbook of $100+\frac{100000}{Q}$, where $Q$ is the number of textbooks produced annually. What should be the annual production volume of the textbook to reach the break-even point if the planned price of the textbook is 300 monetary units?
# | # Solution
At the break-even point $\mathrm{P}=\mathrm{ATC}=\mathrm{MC}$
Form the equation $100+10000 / Q=300$
$100 \mathrm{Q}+100000=300 \mathrm{Q}$
$100000=200 \mathrm{Q}$
$\mathrm{Q}=100000 / 200=500$
## Evaluation Criteria
1. The correct answer is justified: 15 points | 500 | Algebra | olympiads | null | null | \(\boxed{500}\) | # Problem 1. Maximum 15 points
A company that produces educational materials for exam preparation incurs average costs per textbook of $100+\frac{100000}{Q}$, where $Q$ is the number of textbooks produced annually. What should be the annual production volume of the textbook to reach the break-even point if the planned price of the textbook is 300 monetary units?
#
The following text is the beginning part of the answer, which you can refer to for solving the problem:
# Solution
At the break-even point $\mathrm{P}=\mathrm{ATC}=\mathrm{MC}$
Form the equation $100+10000 / Q=300$
$100 \mathrm{Q}+100000=300 |
2. Dima and Seryozha decided to have a competition around a circular lake. They start simultaneously from the same point. Seryozha drives a motorboat at a constant speed of 20 km/h and somehow crosses the lake (not necessarily along the diameter) in 30 minutes. During this time, Dima runs along the shore of the lake for the first 15 minutes at a speed of 6 km/h, and then for the rest of the time, he swims on a boat parallel to Seryozha's trajectory. After arriving simultaneously on the other side of the lake, the boys run towards each other. Throughout the game, the boys run at constant and equal speeds. How long after parting will they meet? | A trapezoid can be inscribed in a circle only if it is isosceles. The length of the segment resting on a chord of the same length in the same circle is the same. Dima runs 1.5 kilometers along the lake. This means they should run 1.5 kilometers together, at a closing speed of 12 km/h, which is 7.5 minutes. In total, 37.5 minutes will have passed from the moment they parted.
| Criteria for Task Evaluation | Points |
| :--- | :---: |
| The correct answer is obtained with justification | 10 |
| The correct answer is obtained, but the solution is either not fully justified or contains an arithmetic error that does not affect the result | 5 |
| The answer is either incorrect or guessed, with no justification provided | 0 | | 37.5 | Algebra | olympiads | null | null | \(\boxed{37.5}\) | 2. Dima and Seryozha decided to have a competition around a circular lake. They start simultaneously from the same point. Seryozha drives a motorboat at a constant speed of 20 km/h and somehow crosses the lake (not necessarily along the diameter) in 30 minutes. During this time, Dima runs along the shore of the lake for the first 15 minutes at a speed of 6 km/h, and then for the rest of the time, he swims on a boat parallel to Seryozha's trajectory. After arriving simultaneously on the other side of the lake, the boys run towards each other. Throughout the game, the boys run at constant and equal speeds. How long after parting will they meet?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A trapezoid can be inscribed in a circle only if it is isosceles. The length of the segment resting on a chord of the same length in the same circle is the same. Dima runs 1.5 kilometers along the lake. This means they should run 1.5 kilometers together, at a closing speed of 12 km/h, which is 7.5 minutes. In total, 37.5 minutes will have passed from the moment |
1. A necklace consists of 30 blue and a certain number of red beads. It is known that on both sides of each blue bead there are beads of different colors, and one bead away from each red bead there are also beads of different colors. How many red beads can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.) | It is obvious that blue beads appear in the necklace in pairs, separated by at least one red bead. Let there be $n$ red beads between two nearest pairs of blue beads. We will prove that $n=4$. Clearly, $n \leqslant 4$, since the middle one of five consecutive red beads does not satisfy the condition of the problem. For $n<4$, there are three possible situations:
CCKKKCC, CCKKCC, CCKCC.
In the first case, the middle red bead does not satisfy the condition, and in the other cases, all red beads do not satisfy the condition.
Thus, pairs of blue beads must be separated by four red beads. It is clear that such a necklace satisfies the condition of the problem. In it, the number of red beads is twice the number of blue beads, that is, there are 60. | 60 | Combinatorics | olympiads | null | null | \(\boxed{60}\) | 1. A necklace consists of 30 blue and a certain number of red beads. It is known that on both sides of each blue bead there are beads of different colors, and one bead away from each red bead there are also beads of different colors. How many red beads can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
It is obvious that blue beads appear in the necklace in pairs, separated by at least one red bead. Let there be $n$ red beads between two nearest pairs of blue beads. We will prove that $n=4$. Clearly, $n \leqslant 4$, since the middle one of five consecutive red beads does not satisfy the condition of the problem. For $n<4$, there are three possible situations:
CCKKKCC, CCK |
1. A string is threaded with 150 beads of red, blue, and green. It is known that among any six consecutive beads, there is at least one green, and among any eleven consecutive beads, there is at least one blue. What is the maximum number of red beads that can be on the string? | We can choose $\left[\frac{150}{11}\right]=13$ consecutive blocks of 11 beads each. Since each block contains at least one blue bead, there are at least 13 blue beads on the string. In addition, we can group all the beads into 25 consecutive blocks of 6 beads each. Each block contains at least one green bead, so there are at least 25 of them on the string. Therefore, the number of red beads is no more than $150-25-13=112$.
Let's provide an example where the string contains exactly 112 red beads. Place the green beads at positions that are multiples of 6, and the blue beads at positions
$$
11,22,33,44,55 ; 65,76,87,98,109 ; 119,130,141
$$
Fill the remaining positions with red beads. | 112 | Combinatorics | olympiads | null | null | \(\boxed{112}\) | 1. A string is threaded with 150 beads of red, blue, and green. It is known that among any six consecutive beads, there is at least one green, and among any eleven consecutive beads, there is at least one blue. What is the maximum number of red beads that can be on the string?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We can choose $\left[\frac{150}{11}\right]=13$ consecutive blocks of 11 beads each. Since each block contains at least one blue bead, there are at least 13 blue beads on the string. In addition, we can group all the beads into 25 consecutive blocks of 6 beads each. Each block contains at least one green bead, so there are at least 25 of them on |
1. A necklace consists of 50 blue and a certain number of red beads. It is known that in any segment of the necklace containing 8 blue beads, there are at least 4 red ones. What is the minimum number of red beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.) | Note that any segment of the necklace consisting of 11 beads contains no more than 7 blue and no fewer than 4 red beads (otherwise, it would contain 8 blue beads and no more than 3 red ones). Fix a red bead in the necklace. The 7 consecutive segments of 11 beads adjacent to it do not cover the entire necklace, as these segments contain no more than 49 blue beads out of 50. Therefore, the total number of red beads is no less than $7 \cdot 4 + 1 = 29$.
Let's provide an example of a necklace containing exactly 29 red beads. Consider the block
$$
\text{B = RC RC RC RC; CCC,}
$$
consisting of 4 red and 7 blue beads. Then the desired necklace has the form
$$
\text{B, B, .., B (7 times); RC.}
$$ | 29 | Combinatorics | olympiads | null | null | \(\boxed{29}\) | 1. A necklace consists of 50 blue and a certain number of red beads. It is known that in any segment of the necklace containing 8 blue beads, there are at least 4 red ones. What is the minimum number of red beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that any segment of the necklace consisting of 11 beads contains no more than 7 blue and no fewer than 4 red beads (otherwise, it would contain 8 blue beads and no more than 3 red ones). Fix a red bead in the necklace. The 7 consecutive segments of 11 beads adjacent to it do not cover the entire necklace, as these segments contain no more than |
1. A necklace consists of 100 red and a certain number of blue beads. It is known that in any segment of the necklace containing 10 red beads, there are at least 7 blue ones. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.) | Note that any segment of the necklace containing 16 beads has no more than 9 red and no fewer than 7 blue beads (otherwise, it would contain 10 red beads and no more than 6 blue ones). Fix a blue bead in the necklace. The 11 consecutive segments of 16 beads adjacent to it do not cover the entire necklace, as these segments contain no more than 99 red beads out of 100. Therefore, the total number of blue beads is no less than $11 \cdot 7+1=78$.
Let's provide an example of a necklace containing exactly 78 blue beads. Consider the block
$$
\text { B = CB CB CB CB CB CB CB; RR, }
$$
consisting of 7 blue and 9 red beads. Then the desired necklace has the form
$$
\text { B, B, .., B (11 times); CR. }
$$ | 78 | Combinatorics | olympiads | null | null | \(\boxed{78}\) | 1. A necklace consists of 100 red and a certain number of blue beads. It is known that in any segment of the necklace containing 10 red beads, there are at least 7 blue ones. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is adjacent to the first.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that any segment of the necklace containing 16 beads has no more than 9 red and no fewer than 7 blue beads (otherwise, it would contain 10 red beads and no more than 6 blue ones). Fix a blue bead in the necklace. The 11 consecutive segments of 16 beads adjacent to it do not cover the entire necklace, as these segments contain no more than 99 red be |
1. A thread is strung with 75 blue, 75 red, and 75 green beads. We will call a sequence of five consecutive beads good if it contains exactly 3 green beads and one each of red and blue. What is the maximum number of good quintets that can be on this thread? | Note that the first and last green beads are included in no more than three good fives, the second and second-to-last - in no more than four fives, and the rest - in no more than five fives. If we add these inequalities, we get $2 \cdot 3 + 2 \cdot 4 + 71 \cdot 5 = 369$ on the right side, and three times the number of good fives on the left side, since each will be counted three times. Therefore, there can be no more than 123 good fives.
Let's provide an example of bead placement that gives exactly 123 good fives:
$$
\text { KSGGG; KSGGG; ...; KSGGG (25 times); KS ... }
$$
(the ellipsis at the end means an arbitrary combination of blue and red beads). The good fives will be those starting at positions $1, 2, \ldots, 123$, and only those. | 123 | Combinatorics | olympiads | null | null | \(\boxed{123}\) | 1. A thread is strung with 75 blue, 75 red, and 75 green beads. We will call a sequence of five consecutive beads good if it contains exactly 3 green beads and one each of red and blue. What is the maximum number of good quintets that can be on this thread?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that the first and last green beads are included in no more than three good fives, the second and second-to-last - in no more than four fives, and the rest - in no more than five fives. If we add these inequalities, we get $2 \cdot 3 + 2 \cdot 4 + 71 \cdot 5 = 369$ on the right side, and three times the number of good fives on the left side, since each will be counted |
1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.) | If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that there are no fewer green beads than red ones. Therefore, there are no fewer than $\frac{80}{3}=26 \frac{2}{3}$ green beads, which means there are at least 27. An example of a necklace containing exactly 27 green beads is as follows:
GKR 3 GKR ... ; GKR (26 times); GK. | 27 | Combinatorics | olympiads | null | null | \(\boxed{27}\) | 1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that the |
1. Around a round table, 50 schoolchildren are sitting: blondes, brunettes, and redheads. It is known that in any group of schoolchildren sitting in a row, between any two blondes there is at least one brunette, and between any two brunettes - at least one redhead. What is the minimum number of redheads that can sit at this table? | If only blondes were sitting at the table, the number of pairs of neighbors would be equal to the number of blondes. Since there is a brunette between any two blondes, there are no fewer brunettes than blondes sitting at the table. Similarly, it can be proven that there are no fewer redheads than brunettes. Therefore, there are no fewer than $\frac{50}{3}=16 \frac{2}{3}$ redheads sitting at the table, which means there are at least 17. Let's provide an example of seating with exactly 17 redheads:
$$
\text { RCB; RCB; ... RCB; (16 times); RC }
$$
(The letters R, C, and B denote redheads, brunettes, and blondes, respectively). | 17 | Combinatorics | olympiads | null | null | \(\boxed{17}\) | 1. Around a round table, 50 schoolchildren are sitting: blondes, brunettes, and redheads. It is known that in any group of schoolchildren sitting in a row, between any two blondes there is at least one brunette, and between any two brunettes - at least one redhead. What is the minimum number of redheads that can sit at this table?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
If only blondes were sitting at the table, the number of pairs of neighbors would be equal to the number of blondes. Since there is a brunette between any two blondes, there are no fewer brunettes than blondes sitting at the table. Similarly, it can be proven that there are no fewer redheads than brunettes. Therefor |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second must be different. For example, 101 and 292 have this form, while 222 and 123 do not. Similarly, we define the form of the number $\overline{a b c a b d}$. How many odd numbers of the form $\overline{\text { adabcd }}$ are divisible by 5? | Odd numbers divisible by 5 are numbers ending in 5, so for $d$ we have only one option. For $a$ we have 8 options, as the number cannot start with zero, and $a$ cannot be equal to $d$. The digit $b$ cannot be equal to $a$ or $d$, and there are no other restrictions on it - we get 8 possible values. Similarly, for the digit $c$ - 7 options. Therefore, the total number of such numbers is $1 \cdot 8 \cdot 8 \cdot 7=448$. | 448 | Combinatorics | olympiads | null | null | \(\boxed{448}\) | 3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second must be different. For example, 101 and 292 have this form, while 222 and 123 do not. Similarly, we define the form of the number $\overline{a b c a b d}$. How many odd numbers of the form $\overline{\text { adabcd }}$ are divisible by 5?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Odd numbers divisible by 5 are numbers ending in 5, so for $d$ we have only one option. For $a$ we have 8 options, as the number cannot start with zero, and $a$ cannot be equal to $d$. The digit $b$ cannot be e |
4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell. | A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will be no fewer than 416 such cells.
We will show how to mark 416 cells in the required manner. Mark all parallel diagonals with lengths of $5, 11, 17, 23, 29, 35, 41$, and $47$. In total, there will be $2 \cdot (5+11+17+23+29+35+41+47)=416$ marked cells. | 416 | Combinatorics | olympiads | null | null | \(\boxed{416}\) | 4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will |
2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression
$$
A=a^{7} b+b^{7} c+c^{7} a+a b^{3}+b c^{3}+c a^{3}
$$ | By the Cauchy-Schwarz inequality
$$
A=\left(a^{7} b+a b^{3}\right)+\left(b^{7} c+b c^{3}\right)+\left(c^{7} a+c a^{3}\right) \geqslant 2\left(a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}\right) \geqslant \frac{2}{3}\left(a^{2} b+b^{2} c+c^{2} a\right)^{2}=6
$$
Equality is achieved when $a=b=c=1$. | 6 | Inequalities | olympiads | null | null | \(\boxed{6}\) | 2. Positive numbers $a, b, c$ are such that $a^{2} b+b^{2} c+c^{2} a=3$. Find the minimum value of the expression
$$
A=a^{7} b+b^{7} c+c^{7} a+a b^{3}+b c^{3}+c a^{3}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By the Cauchy-Schwarz inequality
$$
A=\left(a^{7} b+a b^{3}\right)+\left(b^{7} c+b c^{3}\right)+\left(c^{7} a+c a^{3}\right) \geqslant 2\left(a^ |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)^{3}-26 a b c}
$$ | Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$
(we used the Cauchy inequality). Then
$$
a^{3}+b^{3}+c^{3} \leqslant(a+b+c)^{3}-24 a b c
$$
Let $t=\frac{(a+b+c)^{3}}{a b c}$. By the Cauchy inequality, $t \geqslant 27$, hence
$$
A \leqslant \frac{(a+b+c)^{3}-24 a b c}{(a+b+c)^{3}-26 a b c}=\frac{t-24}{t-26}=1+\frac{2}{t-26} \leqslant 3
$$
Equality is achieved when $a=b=c$. | 3 | Inequalities | olympiads | null | null | \(\boxed{3}\) | 2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{3}+b^{3}+c^{3}}{(a+b+c)^{3}-26 a b c}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+6 a b c+B$, where
$$
B=3\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right) \geqslant 18 \sqrt[6]{a^{6} b^{6} c^{6}}=18 a b c
$$
(we used the Cauchy inequality). Then
$$
a^{3}+b^{3}+c^{3} \leqslant(a+b+c |
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}
$$ | Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where
\[
\begin{aligned}
& B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\
& \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b c)^{4}}+36 \sqrt[3]{(a b c)^{4}}=78(a b c)^{4 / 3}
\end{aligned}
\]
(we used the Cauchy inequality). Then
\[
a^{4}+b^{4}+c^{4} \leqslant(a+b+c)^{4}-78(a b c)^{4 / 3}
\]
Let $t=\frac{(a+b+c)^{4}}{(a b c)^{4 / 3}}$. By the Cauchy inequality, $t \geqslant 81$, hence
\[
A \leqslant \frac{(a+b+c)^{4}-78(a b c)^{4 / 3}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}=\frac{t-78}{t-80}=1+\frac{2}{t-80} \leqslant 3
\]
Equality is achieved when $a=b=c$. | 3 | Inequalities | olympiads | null | null | \(\boxed{3}\) | 2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{4}+b^{4}+c^{4}}{(a+b+c)^{4}-80(a b c)^{4 / 3}}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that $(a+b+c)^{4}=a^{4}+b^{4}+c^{4}+B$, where
\[
\begin{aligned}
& B=4\left(a^{3} b+a b^{3}+b^{3} c+b c^{3}+c^{3} a+c a^{3}\right)+6\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+12\left(a b c^{2}+b c a^{2}+c a b^{2}\right) \geqslant \\
& \geqslant 24 \sqrt[6]{(a b c)^{8}}+18 \sqrt[3]{(a b c)^{4}}+36 \sqrt[3]{(a b c)^{4}}=78(a b c)^{4 / 3}
\end{aligned}
\]
|
2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{a^{3}+b^{3}+c^{3}-2 a b c}
$$ | By the Cauchy-Schwarz inequality,
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)=\frac{1}{3}\left((a+b+c)^{3}-\left(a^{3}+b^{3}+c^{3}+6 a b c\right)\right) \leqslant \frac{1}{3}\left((a+b+c)^{3}-9 a b c\right)
$$
Note that
$$
a^{3}+b^{3}+c^{3}-2 a b c \geqslant \frac{1}{9}(a+b+c)^{3}-2 a b c=\frac{1}{9}\left((a+b+c)^{3}-18 a b c\right)
$$
Let $t=\frac{(a+b+c)^{3}}{a b c}$. By the Cauchy-Schwarz inequality, $t \geqslant 27$, hence
$$
A \leqslant \frac{3\left((a+b+c)^{3}-9 a b c\right)}{(a+b+c)^{3}-18 a b c}=\frac{3(t-9)}{t-18}=3\left(1+\frac{9}{t-18}\right) \leqslant 6
$$
Equality is achieved when $a=b=c$. | 6 | Algebra | olympiads | null | null | \(\boxed{6}\) | 2. Given $a, b, c > 0$, find the maximum value of the expression
$$
A=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{a^{3}+b^{3}+c^{3}-2 a b c}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By the Cauchy-Schwarz inequality,
$$
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)=\frac{1}{3}\left((a+b+c)^{3}-\left(a^{3}+b^{3}+c^{3}+6 a b c\right)\right) \leqslant \frac{1}{3}\left((a+b+c)^{3}-9 a b c\right)
$$
Note that
$$
a^{3}+b^{3}+c^{3}-2 a b c \geqslant \frac{1}{9}(a+b+c)^{3}-2 a b c=\frac{1}{9}\left((a |
1. For what smallest $k$ can $k$ cells be marked on a $10 \times 11$ board such that any placement of a three-cell corner on the board touches at least one marked cell? | It is not hard to notice that in any $2 \times 2$ square, there are at least two marked cells. Since 25 such squares can be cut out from a $10 \times 11$ board, there must be no fewer than 50 marked cells in it. An example with 50 marked cells is obtained if the cells with an even first coordinate are marked. | 50 | Combinatorics | olympiads | null | null | \(\boxed{50}\) | 1. For what smallest $k$ can $k$ cells be marked on a $10 \times 11$ board such that any placement of a three-cell corner on the board touches at least one marked cell?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
It is not hard to notice that in any $2 \times 2$ square, there are at least two marked cells. Since 25 such squares can be cut out from a $10 \times 11$ b |
1. In the black cells of an $8 \times 8$ chessboard, the numbers $1,2,3, \ldots, 32$ are placed such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$. | Divide the chessboard into 16 squares of $2 \times 2$. Since the sum of the numbers in all the black cells is
$$
1+2+\cdots+32=\frac{32 \cdot 33}{2}=16 \cdot 33
$$
the arithmetic mean of the sums of the numbers in these 16 squares is 33. Therefore, in at least one square, the sum of the numbers is not less than 33, i.e., $S \geqslant 33$. An example of an arrangement where $S=33$ is achieved is shown in the figure.
| 32 | | 31 | | 30 | | 29 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 1 | | 2 | | 3 | | 4 |
| 28 | | 27 | | 26 | | 25 | |
| | 5 | | 6 | | 7 | | 8 |
| 24 | | 23 | | 22 | | 21 | |
| | 9 | | 10 | | 11 | | 12 |
| 20 | | 19 | | 18 | | 17 | |
| | 13 | | 14 | | 15 | | 16 | | 33 | Combinatorics | olympiads | null | null | \(\boxed{33}\) | 1. In the black cells of an $8 \times 8$ chessboard, the numbers $1,2,3, \ldots, 32$ are placed such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Divide the chessboard into 16 squares of $2 \times 2$. Since the sum of the numbers in all the black cells is
$$
1+2+\cdots+32=\frac{32 \cdot 33}{2}=16 \cdot 33
$$
the arithmetic mean of the sums of the numbers in these 16 squares is 33. Therefore, in at least one square, the sum of the numbers is not less than 33, i.e., $S \geqslant 33$. An example of an arrangement whe |
4. In a box, there is a large batch of flowers of six types mixed together. Vasya randomly takes flowers one by one from the box. As soon as he collects 5 flowers of the same type, he makes a bouquet and sells it. What is the minimum number of flowers he needs to take to guarantee selling 10 bouquets? | Note that 69 flowers are not enough. Indeed, if Vasya pulled out 49 flowers of the first type and 4 flowers of each of the other types, then in total he took $49+5 \cdot 4=69$ flowers, but from them, he can only make 9 bouquets.
Suppose Vasya pulled out the 70th flower and possibly sold one bouquet. From the remaining flowers, it is impossible to form another bouquet. Therefore, Vasya has no more than four flowers of each type, and in total,
no more than 24 flowers. Since both 70 and the number of sold flowers are multiples of 5, no more than 20 flowers could remain. This means Vasya sold at least 50 flowers, that is, 10 bouquets. | 70 | Combinatorics | olympiads | null | null | \(\boxed{70}\) | 4. In a box, there is a large batch of flowers of six types mixed together. Vasya randomly takes flowers one by one from the box. As soon as he collects 5 flowers of the same type, he makes a bouquet and sells it. What is the minimum number of flowers he needs to take to guarantee selling 10 bouquets?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that 69 flowers are not enough. Indeed, if Vasya pulled out 49 flowers of the first type and 4 flowers of each of the other types, then in total he took $49+5 \cdot 4=69$ flowers, but from them, he can only make 9 bouquets.
Suppose Vasya pulled out the 70th flower and possibly sold one bouquet. From the remaining |
4. What is the maximum number of vertices of a regular 2016-gon that can be marked so that no four marked vertices are the vertices of any rectangle? | Note that an inscribed quadrilateral is a rectangle if and only if its diagonals are diameters of the circumscribed circle. The 2016-gon has exactly 1008 pairs of diametrically opposite vertices. If no rectangle can be formed from the marked vertices, then only in one pair can both vertices be marked. Therefore, no more than \(1007 + 2 = 1009\) vertices can be marked.
On the other hand, number the vertices of the 2016-gon in order from 1 to 2016 and mark the first 1009 vertices. Only the vertices numbered 1 and 1009 will be diametrically opposite. Therefore, 1009 satisfies the condition of the problem. | 1009 | Combinatorics | olympiads | null | null | \(\boxed{1009}\) | 4. What is the maximum number of vertices of a regular 2016-gon that can be marked so that no four marked vertices are the vertices of any rectangle?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that an inscribed quadrilateral is a rectangle if and only if its diagonals are diameters of the circumscribed circle. The 2016-gon has exactly 1008 pairs of diametrically opposite vertices. If no rectangle can be formed from the marked vertices, then only in one pair can both vertices be marked. Th |
2. (10 points) Minister K. issued an order that citizens will only be received if the number of ways to choose a group of four people from those who have come is less than the number of ways to choose a group of two people from them. Determine the maximum number of citizens that the minister can receive? | The number of ways to choose a group of four people from $n$ people is $C_{n}^{4}$; a group of two people is $C_{n}^{2}$. We are interested in the maximum natural number $n$ such that $C_{n}^{4}<C_{n}^{2}$. Transforming:
$$
\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4}<\frac{n(n-1)}{2} \Leftrightarrow \frac{(n-2)(n-3)}{3 \cdot 4}<1
$$
Solving the inequality using the interval method, we get that $-1<n<6$. Therefore, the maximum $n$ is 5. | 5 | Combinatorics | olympiads | null | null | \(\boxed{5}\) | 2. (10 points) Minister K. issued an order that citizens will only be received if the number of ways to choose a group of four people from those who have come is less than the number of ways to choose a group of two people from them. Determine the maximum number of citizens that the minister can receive?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number of ways to choose a group of four people from $n$ people is $C_{n}^{4}$; a group of two people is $C_{n}^{2}$. We are interested in the maximum natural number $n$ such that $C_{n}^{4}<C_{n}^{2}$. Transforming:
$$ |
3. (20 points) Young marketer Masha was supposed to survey 50 customers in an electronics store throughout the day. However, there were fewer customers in the store that day. What is the maximum number of customers Masha could have surveyed, given that according to her data, 7 of the respondents made an impulse purchase, 75% of the remaining respondents bought the product under the influence of advertising, and the number of customers who chose the product based on the advice of a sales consultant is one-third of the number who chose the product under the influence of advertising. | Let the number of customers surveyed be $x$. Then, the number of customers who made a purchase under the influence of advertising is $(x-7) \cdot 3 / 4$, and the number of customers who made a purchase on the advice of a sales consultant is $(x-7)/4$. Since the number of customers can only be an integer, $x-7$ must be divisible by 4. The maximum suitable number $x$, less than 50, is 47. | 47 | Algebra | olympiads | null | null | \(\boxed{47}\) | 3. (20 points) Young marketer Masha was supposed to survey 50 customers in an electronics store throughout the day. However, there were fewer customers in the store that day. What is the maximum number of customers Masha could have surveyed, given that according to her data, 7 of the respondents made an impulse purchase, 75% of the remaining respondents bought the product under the influence of advertising, and the number of customers who chose the product based on the advice of a sales consultant is one-third of the number who chose the product under the influence of advertising.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the number of customers surveyed be $x$. Then, the number of customers who made a purchase under the influence of advertising is $(x-7) \cdot 3 / 4$, and the number of customers who made a p |
4. (20 points) In the Martian calendar, a year consists of 5882 days, and each month has either 100 or 77 days. How many months are there in the Martian calendar? | Let $x$ be the number of months with 100 days, and $y$ be the number of months with 77 days. According to the problem, $100 x + 77 y = 5882$. It is obvious that $y \leqslant 66$, otherwise $x < 0$. Notice that
$$
x \bmod 11 = 100 x \quad \bmod 11 = 5882 \quad \bmod 11 = 8
$$
Thus, $x = 11 k + 8$ for some integer $k$, from which we have
$$
(11 k + 8) \cdot 100 + 77 y = 5882 \Longleftrightarrow 1100 k = 5082 - 77 y = 77(66 - y)
$$
The right side is divisible by 100, but 77 and 100 are coprime. Therefore, the number $66 - y$, which lies between 0 and 65, must be divisible by 100. This is only possible when $y = 66$. Therefore, $x = 0.01 \cdot (5882 - 76 \cdot 66) = 8$ and $x + y = 74$. | 74 | Number Theory | olympiads | null | null | \(\boxed{74}\) | 4. (20 points) In the Martian calendar, a year consists of 5882 days, and each month has either 100 or 77 days. How many months are there in the Martian calendar?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $x$ be the number of months with 100 days, and $y$ be the number of months with 77 days. According to the problem, $100 x + 77 y = 5882$. It is obvious that $y \leqslant 66$, otherwise $x < 0$. Notice that
$$
x \bmod 11 = 100 x \quad \bmod 11 = 5882 \quad \bmod 11 = 8
$$
Thus, $x = 11 k + 8$ for some integer $k$, from which we have
$$
(11 |
5. (20 points) Professor K., wishing to be known as a wit, plans to tell no fewer than two but no more than three different jokes at each of his lectures. At the same time, the sets of jokes told at different lectures should not coincide. How many lectures in total will Professor K. be able to give if he knows 8 jokes? | The professor can use all possible triplets of anecdotes (the number of which is $C_{8}^{3}=56$), as well as all pairs of anecdotes (which is $C_{8}^{2}=28$). Therefore, the maximum number of sets of anecdotes the professor can use in lectures is $56+28=84$. | 84 | Combinatorics | olympiads | null | null | \(\boxed{84}\) | 5. (20 points) Professor K., wishing to be known as a wit, plans to tell no fewer than two but no more than three different jokes at each of his lectures. At the same time, the sets of jokes told at different lectures should not coincide. How many lectures in total will Professor K. be able to give if he knows 8 jokes?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The professor can use all possible triplets of anecdotes (the number of which is $C_{8}^{3}=56$), as well as all pairs of anecdot |
6. (20 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different five-letter words can be formed from the letters of the word САМСА? And from the letters of the word ПАСТА? In your answer, indicate the sum of the found numbers. | In the word SAMSA, the letter A appears twice and the letter S appears twice. Therefore, the number of different words will be $\frac{5!}{2!\cdot 2!}=30$. In the word PASTA, only the letter A appears twice. Therefore, the number of different words in this case will be $\frac{5!}{2!}=60$. In total, we get $30+60=90$. | 90 | Combinatorics | olympiads | null | null | \(\boxed{90}\) | 6. (20 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different five-letter words can be formed from the letters of the word САМСА? And from the letters of the word ПАСТА? In your answer, indicate the sum of the found numbers.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In the word SAMSA, the letter A appears twice and the letter S appears twice. Therefore, the number of different words will be $\frac{5!}{2!\cdot 2!}=30$. In |
9. (40 points) On an island, there live only 50 knights, who always tell the truth, and 15 commoners, who can either tell the truth or lie. A scatterbrained professor, who came to the island to give a lecture, forgot what color hat he was wearing. How many of the local residents should the professor ask about the color of his hat to be sure of what it is? | Since the professor will only ask about the color of the hat, to accurately determine the color of the hat, it is necessary to survey more knights than commoners. In the worst case, among those surveyed, there could be all the commoners on the island, i.e., 15 people; therefore, it is necessary to survey no fewer than 31 people to ensure that the opinion of the knights prevails. | 31 | Logic and Puzzles | olympiads | null | null | \(\boxed{31}\) | 9. (40 points) On an island, there live only 50 knights, who always tell the truth, and 15 commoners, who can either tell the truth or lie. A scatterbrained professor, who came to the island to give a lecture, forgot what color hat he was wearing. How many of the local residents should the professor ask about the color of his hat to be sure of what it is?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since the professor will only ask about the color of the hat, to accurately determine the color of the hat, it is necessary to survey more knights than commoners. In the worst case, among th |
3. (20 points) During the draw before the math marathon, team captains were asked to name the smallest possible sum of the digits in the decimal representation of the number $n+1$, given that the sum of the digits of the number $n$ is 2017. What was the answer given by the captain of the team that won the draw? | First, we show that the answer is not less than 2. If the sum of the digits of the number $n+1$ is 1, then $n+1=10 \ldots 0$, and the decimal representation of $n$ consists entirely of nines. Then the number $n$ is divisible by 9, and the sum of its digits, therefore, is also. But this is impossible since 2017 is not divisible by 9.
It remains to provide an example of a number $n+1$ whose sum of digits is 2. Since $2017=224 \cdot 9+1$, we can set $n=19 \ldots 9$, where the digit 9 is repeated 224 times. Indeed, in this case, $n+1=20 \ldots 0$. | 2 | Number Theory | olympiads | null | null | \(\boxed{2}\) | 3. (20 points) During the draw before the math marathon, team captains were asked to name the smallest possible sum of the digits in the decimal representation of the number $n+1$, given that the sum of the digits of the number $n$ is 2017. What was the answer given by the captain of the team that won the draw?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
First, we show that the answer is not less than 2. If the sum of the digits of the number $n+1$ is 1, then $n+1=10 \ldots 0$, and the decimal representation of $n$ consists entirely of nines. Then the number $n$ is divisible by 9, and the sum of its digits, therefore, is als |
4. (20 points) Before the math lesson, the teacher wrote nine consecutive numbers on the board, but the duty students accidentally erased one of them. When the lesson began, it turned out that the sum of the remaining eight numbers is 1703. Which number did the duty students erase? | Let the average of the original numbers be $a$. Then these numbers can be written in a symmetric form:
$$
a-4, a-3, a-2, a-1, a, a+1, a+2, a+3, a+4
$$
The erased number has the form $a+b$, where $-4 \leqslant b \leqslant 4$, and the sum of the remaining numbers is $9a - (a+b) = 8a - b$. On the other hand, this sum is also equal to $1703 = 8 \cdot 213 - 1$. Therefore,
$$
8 \cdot 213 - 1 = 8a - b, \quad \text{from which} \quad 8(a - 213) = b - 1
$$
Then the number $b-1$ is divisible by 8 and lies in the interval $[-5, 3]$. Thus, $b=1, a=213$, and the number that was erased is $213 + 1 = 214$. | 214 | Number Theory | olympiads | null | null | \(\boxed{214}\) | 4. (20 points) Before the math lesson, the teacher wrote nine consecutive numbers on the board, but the duty students accidentally erased one of them. When the lesson began, it turned out that the sum of the remaining eight numbers is 1703. Which number did the duty students erase?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the average of the original numbers be $a$. Then these numbers can be written in a symmetric form:
$$
a-4, a-3, a-2, a-1, a, a+1, a+2, a+3, a+4
$$
The erased number has the form $a+b$, where $-4 \leqslant b \leqslant 4$, and the sum of the remaining numbers is $9a - (a+b) = 8a - b$. On the oth |
2. For $x, y, z \in (0,2]$, find the maximum value of the expression
$$
A=\frac{\left(x^{3}-6\right) \sqrt[3]{x+6}+\left(y^{3}-6\right) \sqrt[3]{y+6}+\left(z^{3}-6\right) \sqrt[3]{z+6}}{x^{2}+y^{2}+z^{2}}
$$ | For $x \in(0,2]$, the inequalities $\sqrt[3]{x+6} \leqslant 2$ and $x^{3} \leqslant 2 x^{2}$ hold, from which
$$
\left(x^{3}-6\right) \sqrt[3]{x+6} \leqslant 2\left(2 x^{2}-6\right)
$$
Similarly, the other two terms in the numerator of $A$ are estimated. Therefore,
$$
A \leqslant 2 \cdot \frac{2 x^{2}-6+2 y^{2}-6+2 z^{2}-6}{x^{2}+y^{2}+z^{2}}=4-\frac{36}{x^{2}+y^{2}+z^{2}} \leqslant 4-\frac{36}{12}=1
$$
Equality is achieved when $x=y=z=2$. | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | 2. For $x, y, z \in (0,2]$, find the maximum value of the expression
$$
A=\frac{\left(x^{3}-6\right) \sqrt[3]{x+6}+\left(y^{3}-6\right) \sqrt[3]{y+6}+\left(z^{3}-6\right) \sqrt[3]{z+6}}{x^{2}+y^{2}+z^{2}}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
For $x \in(0,2]$, the inequalities $\sqrt[3]{x+6} \leqslant 2$ and $x^{3} \leqslant 2 x^{2}$ hold, from which
$$
\left(x^{3}-6\right) \sqrt[3]{x+6} \leqslant 2\left(2 x^{2}-6\right)
$$
Similarly, the other two terms in the |
2. Given $x, y, z \geqslant 3$ find the minimum value of the expression
$$
A=\frac{\left(x^{3}-24\right) \sqrt[3]{x+24}+\left(y^{3}-24\right) \sqrt[3]{y+24}+\left(z^{3}-24\right) \sqrt[3]{z+24}}{x y+y z+z x}
$$ | For $x \geqslant 3$, the inequalities $\sqrt[3]{x+24} \geqslant 3$ and $x^{3} \geqslant 3 x^{2}$ hold, from which
$$
\left(x^{3}-24\right) \sqrt[3]{x+24} \geqslant 3\left(3 x^{2}-24\right)
$$
Similarly, the other two terms in the numerator of $A$ are estimated. Therefore,
$$
A \geqslant 3 \cdot \frac{3 x^{2}-24+3 y^{2}-24+3 z^{2}-24}{x y+y z+z x}=9\left(\frac{x^{2}+y^{2}+z^{2}}{x y+y z+z x}-\frac{24}{x y+y z+z x}\right) \geqslant 9\left(1-\frac{24}{27}\right)=1
$$
Equality is achieved when $x=y=z=3$. | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | 2. Given $x, y, z \geqslant 3$ find the minimum value of the expression
$$
A=\frac{\left(x^{3}-24\right) \sqrt[3]{x+24}+\left(y^{3}-24\right) \sqrt[3]{y+24}+\left(z^{3}-24\right) \sqrt[3]{z+24}}{x y+y z+z x}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
For $x \geqslant 3$, the inequalities $\sqrt[3]{x+24} \geqslant 3$ and $x^{3} \geqslant 3 x^{2}$ hold, from which
$$
\left(x^{3}-24\right) \sqrt[3]{x+24} \geqslant 3\left(3 x^{2}-24\right)
$$
Similarly, the other two terms in the numerator of $A$ are e |
2. Given $x, y, z \geqslant 1$, find the minimum value of the expression
$$
A=\frac{\sqrt{3 x^{4}+y}+\sqrt{3 y^{4}+z}+\sqrt{3 z^{4}+x}-3}{x y+y z+z x}
$$ | Note that for $x, y \geqslant 1$
$$
\sqrt{3 x^{4}+y}-1=\sqrt{x^{4}+2 x^{4}+y}-1 \geqslant \sqrt{x^{4}+2 x^{2}+1}-1=x^{2}
$$
Similarly, it can be shown that
$$
\sqrt{3 y^{4}+z}-1 \geqslant y^{2}, \quad \sqrt{3 z^{4}+x}-1 \geqslant z^{2}
$$
Therefore,
$$
A \geqslant \frac{x^{2}+y^{2}+z^{2}}{x y+y z+z x} \geqslant 1
$$
Equality is achieved when $x=y=z=1$. | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | 2. Given $x, y, z \geqslant 1$, find the minimum value of the expression
$$
A=\frac{\sqrt{3 x^{4}+y}+\sqrt{3 y^{4}+z}+\sqrt{3 z^{4}+x}-3}{x y+y z+z x}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that for $x, y \geqslant 1$
$$
\sqrt{3 x^{4}+y}-1=\sqrt{x^{4}+2 x^{4}+y}-1 \geqslant \sqrt{x^{4}+2 x^{2}+1}-1=x^{2}
$$
Similarly, it can be shown that
$$
\sqrt{3 y^{4}+z}-1 |
2. For $x, y, z \in(0,1]$ find the maximum value of the expression
$$
A=\frac{\sqrt{8 x^{4}+y}+\sqrt{8 y^{4}+z}+\sqrt{8 z^{4}+x}-3}{x+y+z}
$$ | Note that for $x, y \in(0,1]$
$$
\sqrt{8 x^{4}+y}-1=\sqrt{4 x^{4}+4 x^{4}+y}-1 \leqslant \sqrt{4 x^{2}+4 x+1}-1=2 x .
$$
Similarly, it can be shown that
$$
\sqrt{3 y^{4}+z}-1 \leqslant 2 y, \quad \sqrt{3 z^{4}+x}-1 \leqslant 2 z
$$
Therefore,
$$
A \leqslant \frac{2 x+2 y+2 z}{x+y+z}=2
$$
Equality is achieved when $x=y=z=1$. | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 2. For $x, y, z \in(0,1]$ find the maximum value of the expression
$$
A=\frac{\sqrt{8 x^{4}+y}+\sqrt{8 y^{4}+z}+\sqrt{8 z^{4}+x}-3}{x+y+z}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that for $x, y \in(0,1]$
$$
\sqrt{8 x^{4}+y}-1=\sqrt{4 x^{4}+4 x^{4}+y}-1 \leqslant \sqrt{4 x^{2}+4 x+1}-1=2 x .
$$
Similarly, it can be shown that
$$
\sqrt{3 |
2. For $x, y \in(0,1]$, find the maximum value of the expression
$$
A=\frac{\left(x^{2}-y\right) \sqrt{y+x^{3}-x y}+\left(y^{2}-x\right) \sqrt{x+y^{3}-x y}+1}{(x-y)^{2}+1}
$$ | Note that
$$
y+x^{3}-x y-x^{2}=\left(y-x^{2}\right)(1-x)
$$
If $y \geqslant x^{2}$, then
$$
\sqrt{y+x^{3}-x y} \geqslant x \quad \text { and } \quad\left(x^{2}-y\right)\left(\sqrt{y+x^{3}-x y}-x\right) \leqslant 0
$$
and when $y<x^{2}$
$$
\sqrt{y+x^{3}-x y} \leqslant x \quad \text { and } \quad\left(x^{2}-y\right)\left(\sqrt{y+x^{3}-x y}-x\right) \leqslant 0
$$
In both cases, the inequality $\left(x^{2}-y\right) \sqrt{y+x^{3}-x y} \leqslant x^{3}-x y$ holds. Estimating the second term in the numerator of $A$ similarly, we get
$$
A \leqslant \frac{x^{3}+y^{3}-2 x y+1}{(x-y)^{2}+1} \leqslant \frac{x^{2}+y^{2}-2 x y+1}{(x-y)^{2}+1}=1
$$
Equality is achieved when $x=y=1$. | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | 2. For $x, y \in(0,1]$, find the maximum value of the expression
$$
A=\frac{\left(x^{2}-y\right) \sqrt{y+x^{3}-x y}+\left(y^{2}-x\right) \sqrt{x+y^{3}-x y}+1}{(x-y)^{2}+1}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that
$$
y+x^{3}-x y-x^{2}=\left(y-x^{2}\right)(1-x)
$$
If $y \geqslant x^{2}$, then
$$
\sqrt{y+x^{3}-x y} \geqslant x \quad \text { and } \quad\left(x^{2}-y\right)\left(\sqrt{y+x^{3}-x y}-x\right) \leqslant 0
$$
and when $y<x^{2}$
$$
\sqrt{y+x^{3}-x y} \leqslant x \quad \text { and } \quad\left(x^{2}-y\right)\left(\sqrt{y+x^{3}-x |
2. For $x, y \in[1,3]$ find the minimum value of the expression
$$
A=\frac{\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y}+\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x}}{x^{2} y+y^{2} x}
$$ | Note that
$$
3 x y+x-3 y-x^{2}=(3 y-x)(x-1) \geqslant 0
$$
since $x \geqslant 1$ and $x \leqslant 3 \leqslant 3 y$. Then
$$
\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y} \geqslant 3 x^{2} y+x^{3} \quad \text { and, similarly, } \quad\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x} \geqslant 3 y^{2} x+y^{3} .
$$
Moreover, by the Cauchy inequality
$$
x^{2} y+y^{2} x=x y(x+y) \leqslant \frac{1}{4}(x+y)^{3}
$$
Therefore,
$$
A \geqslant \frac{3 x y(x+y)+x^{3}+y^{3}}{\frac{1}{4}(x+y)^{3}}=4 \cdot \frac{(x+y)^{3}}{(x+y)^{3}}=4
$$
Equality is achieved when $x=y=1$. | 4 | Algebra | olympiads | null | null | \(\boxed{4}\) | 2. For $x, y \in[1,3]$ find the minimum value of the expression
$$
A=\frac{\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y}+\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x}}{x^{2} y+y^{2} x}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that
$$
3 x y+x-3 y-x^{2}=(3 y-x)(x-1) \geqslant 0
$$
since $x \geqslant 1$ and $x \leqslant 3 \leqslant 3 y$. Then
$$
\left(3 x y+x^{2}\right) \sqrt{3 x y+x-3 y} \geqslant 3 x^{2} y+x^{3} \quad \text { and, similarly, } \quad\left(3 x y+y^{2}\right) \sqrt{3 x y+y-3 x} \geq |
2. For $x, y, z \in(0,1]$ find the minimum value of the expression
$$
A=\frac{(x+2 y) \sqrt{x+y-x y}+(y+2 z) \sqrt{y+z-y z}+(z+2 x) \sqrt{z+x-z x}}{x y+y z+z x}
$$ | Note that for $x, y \in(0,1]$
$$
\sqrt{x+y-x y}=\sqrt{x+y(1-x)} \geqslant \sqrt{x} \geqslant x
$$
Then $(x+2 y) \sqrt{x+y-x y} \geqslant x^{2}+2 x y$ and similarly,
$$
(y+2 z) \sqrt{y+z-y z} \geqslant y^{2}+2 y z, \quad(z+2 x) \sqrt{z+x-z x} \geqslant z^{2}+2 z x
$$
Therefore,
$$
A \geqslant \frac{x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x}{x y+y z+z x}=\frac{(x+y+z)^{2}}{x y+y z+z x} \geqslant 3
$$
Equality is achieved when $x=y=z=1$. | 3 | Algebra | olympiads | null | null | \(\boxed{3}\) | 2. For $x, y, z \in(0,1]$ find the minimum value of the expression
$$
A=\frac{(x+2 y) \sqrt{x+y-x y}+(y+2 z) \sqrt{y+z-y z}+(z+2 x) \sqrt{z+x-z x}}{x y+y z+z x}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that for $x, y \in(0,1]$
$$
\sqrt{x+y-x y}=\sqrt{x+y(1-x)} \geqslant \sqrt{x} \geqslant x
$$
Then $(x+2 y) \sqrt{x+y-x y} \geqslant x^{2}+2 x y$ and similarly,
$$
(y+2 z) \sqrt{y+z-y z} \geqslant y^{2}+2 y z, \q |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers, the difference of whose square roots is less than 1? | We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:
\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]
Since there are 45 numbers, some two of them (let's call them $a$ and $b$) will end up in the same group. Suppose for definiteness that $a<b$. Then $k^{2} \leqslant a<b<(k+1)^{2}$ and, consequently, $\sqrt{b}-\sqrt{a}<(k+1)-k=1$.
Now let's present 44 numbers, all differences between the square roots of which are at least 1:
\[
\begin{array}{llllll}
1^{2}, & 2^{2}, & 3^{2}, & 4^{2}, & \ldots, & 44^{2} .
\end{array}
\] | 45 | Number Theory | olympiads | null | null | \(\boxed{45}\) | 2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers, the difference of whose square roots is less than 1?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:
\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]
Since there are 45 numbers, some two of them (let's call them $a$ and $ |
2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6. | Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then $c-b=6$, which is impossible. Therefore, in each set of ten, no more than four numbers are selected, and in the first thousand numbers, there are no more than 400, since there are a hundred sets of ten in a thousand.
If we take all the numbers ending in $1, 2, 3$ or 4, then there will be exactly 400, but no difference will be equal to 4, 5 or 6. | 400 | Combinatorics | olympiads | null | null | \(\boxed{400}\) | 2. What is the maximum number of different numbers from 1 to 1000 that can be chosen so that the difference between any two chosen numbers is not equal to any of the numbers 4, 5, 6.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Consider ten consecutive natural numbers. We will prove that no more than four of them are selected. If at least five numbers are selected, then three of them have the same parity, but then their pairwise differences cannot be only 2 and 8. Indeed, if $a<b<c$, then $b-a=2$ and $c-a=8$, but then $c-b=6$, which is impossib |
1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-4 f(x)=0$ has exactly three solutions. How many solutions does the equation $(f(x))^{2}=1$ have? | Suppose the leading coefficient of the polynomial is positive. Note that $(f(x))^{3}-4 f(x)=f(x) \cdot(f(x)-2) \cdot(f(x)+2)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-2$, and fewer roots than the equation $f(x)=2$. It is also clear that no two equations have common roots. Then the equation $f(x)=0$ has exactly one root. Therefore, the equation $f(x)=1$ has exactly two roots, and the equation $f(x)=-1$ has no roots. Thus, the equation $(f(x))^{2}-1=(f(x)+1)(f(x)-1)=0$ has two roots. Similarly, the case when the leading coefficient of the polynomial is negative is considered. | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-4 f(x)=0$ has exactly three solutions. How many solutions does the equation $(f(x))^{2}=1$ have?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Suppose the leading coefficient of the polynomial is positive. Note that $(f(x))^{3}-4 f(x)=f(x) \cdot(f(x)-2) \cdot(f(x)+2)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-2$, and fewer roots than the equation $f(x)=2$. It is also clear that no two equations have common roots. Then t |
2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$? | We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:
\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]
Since there are 13 numbers, some two of them (let's call them $a$ and $b$) will end up in the same group. Let's assume for definiteness that $a<b$. Then $k^{3} \leqslant a<b<(k+1)^{3}$ and, consequently, $0<\sqrt[3]{b}-\sqrt[3]{a}<(k+1)-k=1$.
Now let's present 12 numbers, all differences between the cube roots of which are at least 1:
\[
1^{3}, \quad 2^{3}, \quad 3^{3}, \quad 4^{3}, \quad \ldots, \quad 12^{3} .
\] | 13 | Number Theory | olympiads | null | null | \(\boxed{13}\) | 2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:
\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]
Since there are 13 numbers, some two of them (let's call them $a$ an |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie. One fine day, 30 islanders sat around a round table. Each of them can see everyone except themselves and their neighbors. Each person in turn said the phrase: "All I see are liars." How many liars were sitting at the table? | Not all of those sitting at the table are liars (otherwise they would all be telling the truth). Therefore, there is at least one knight sitting at the table. Everyone he sees is a liar. Let's determine who his neighbors are. Both of them cannot be liars (otherwise they would be telling the truth). Also, both of them cannot be knights, since they see each other. Therefore, one of them is a knight, and the other is a liar. It remains to note that the situation where two knights sit next to each other, and all the others are liars, is possible. | 28 | Logic and Puzzles | olympiads | null | null | \(\boxed{28}\) | 1. On an island, there live only knights, who always tell the truth, and liars, who always lie. One fine day, 30 islanders sat around a round table. Each of them can see everyone except themselves and their neighbors. Each person in turn said the phrase: "All I see are liars." How many liars were sitting at the table?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Not all of those sitting at the table are liars (otherwise they would all be telling the truth). Therefore, there is at least one knight sitting at the table. Everyone he sees is a liar. Let's determine who his neighbors are. Both of them cannot be liars (otherwise they wou |
1. On an island, there live only knights, who always tell the truth, and liars, who always lie, and there are at least two knights and at least two liars. One fine day, each islander, in turn, pointed to each of the others and said one of two phrases: "You are a knight!" or "You are a liar!" The phrase "You are a liar!" was heard exactly 230 times. How many times was the phrase "You are a knight!" heard? | Let $r$ and $\ell$ denote the number of knights and liars, respectively. Note that a knight will say to another knight and a liar will say to another liar: "You are a knight!", while a knight will say to a liar and a liar will say to a knight: "You are a liar!" Therefore, the number of liar-knight pairs is $\frac{230}{2}=115=r \ell$. Since $r \ell=115=5 \cdot 23$ and $r, \ell \geqslant 2$, either $r=5$ and $\ell=23$, or $r=23$ and $\ell=5$. In either case, the number of knight-knight and liar-liar pairs is $\frac{5 \cdot 4}{2}+\frac{23 \cdot 22}{2}=263$. Therefore, the phrase "You are a knight!" was said 526 times. | 526 | Logic and Puzzles | olympiads | null | null | \(\boxed{526}\) | 1. On an island, there live only knights, who always tell the truth, and liars, who always lie, and there are at least two knights and at least two liars. One fine day, each islander, in turn, pointed to each of the others and said one of two phrases: "You are a knight!" or "You are a liar!" The phrase "You are a liar!" was heard exactly 230 times. How many times was the phrase "You are a knight!" heard?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $r$ and $\ell$ denote the number of knights and liars, respectively. Note that a knight will say to another knight and a liar will say to another liar: "You are a knight!", while a knight will say to a liar and a liar will say to a knight: "You are a liar!" Therefore, the number of liar-knight pairs is $\f |
3. Real numbers $a, b, c$ and $d$ satisfy the condition $a^{6}+b^{6}+c^{6}+d^{6}=64$. Find the maximum value of the expression $a^{7}+b^{7}+c^{7}+d^{7}$. | By the condition $a^{6} \leqslant a^{6}+b^{6}+c^{6}+d^{6}=64$, therefore $a \leqslant 2$. Similarly, we get that $b \leqslant 2, c \leqslant 2$ and $d \leqslant 2$. Consequently,
$$
a^{7}+b^{7}+c^{7}+d^{7}=a \cdot a^{6}+b \cdot b^{6}+c \cdot c^{6}+d \cdot d^{6} \leqslant 2\left(a^{6}+b^{6}+c^{6}+d^{6}\right)=2 \cdot 64=128
$$
Equality is achieved when $a=2, b=c=d=0$. Therefore, $a^{7}+b^{7}+c^{7}+d^{7}$ does not exceed 128 and can be equal to it. | 128 | Algebra | olympiads | null | null | \(\boxed{128}\) | 3. Real numbers $a, b, c$ and $d$ satisfy the condition $a^{6}+b^{6}+c^{6}+d^{6}=64$. Find the maximum value of the expression $a^{7}+b^{7}+c^{7}+d^{7}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By the condition $a^{6} \leqslant a^{6}+b^{6}+c^{6}+d^{6}=64$, therefore $a \leqslant 2$. Similarly, we get that $b \leqslant 2, c \leqslant 2$ and $d \leqslant 2$. Consequently,
$$
a^{7}+b^{7}+c^{7}+d^{7}=a \cdot a^{6}+b \cd |
1. (15 points) If all the trees on one hectare of forest are cut down, then 100 cubic meters of boards can be produced. Assuming that all the trees in the forest are the same, are evenly distributed, and that 0.4 m$^{3}$ of boards can be obtained from each tree, determine the area in square meters on which one tree grows. In your answer, write the integer part of the obtained number. (The integer part of a number is the greatest integer not exceeding the given number.) | Let's find out how many trees grow on one hectare of forest: $\frac{100 \mathrm{m}^{3}}{0.4 \mathrm{M}^{3}}=250$. Let's recall that 1 ha is $100 \mathrm{~m} \times 100 \mathrm{~m}=10000 \mathrm{~m}^{2}$. Thus, one tree grows on $\frac{10000 \mathrm{M}^{2}}{250}=40 \mathrm{M}^{2}$. | 40 | Algebra | olympiads | null | null | \(\boxed{40}\) | 1. (15 points) If all the trees on one hectare of forest are cut down, then 100 cubic meters of boards can be produced. Assuming that all the trees in the forest are the same, are evenly distributed, and that 0.4 m$^{3}$ of boards can be obtained from each tree, determine the area in square meters on which one tree grows. In your answer, write the integer part of the obtained number. (The integer part of a number is the greatest integer not exceeding the given number.)
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's find out how many trees grow on one hectare of forest: $\frac{100 \mathrm{m}^{3}}{0.4 \mathrm{M}^{3}}=250$. Let's recall that 1 ha is |
2. (15 points) Witch Gingema enchanted the wall clock so that the minute hand moves in the correct direction for five minutes, then three minutes in the opposite direction, then again five minutes in the correct direction, and so on. How many minutes will the hand show after 2022 minutes from the moment it pointed exactly at 12 o'clock before the start of the five-minute correct movement interval? | In 8 minutes of magical time, the hand will move 2 minutes in the clockwise direction. Therefore, in 2022 minutes, it will complete 252 full eight-minute cycles and have 6 minutes left. Since $252 \cdot 2=60 \cdot 8+24$, the hand will travel 8 full circles, plus 24 minutes, and then 5 minutes in the correct direction and 1 minute in the opposite direction. In total, $24+5-1=28$. | 28 | Logic and Puzzles | olympiads | null | null | \(\boxed{28}\) | 2. (15 points) Witch Gingema enchanted the wall clock so that the minute hand moves in the correct direction for five minutes, then three minutes in the opposite direction, then again five minutes in the correct direction, and so on. How many minutes will the hand show after 2022 minutes from the moment it pointed exactly at 12 o'clock before the start of the five-minute correct movement interval?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In 8 minutes of magical time, the hand will move 2 minutes in the clockwise direction. Therefore, in 2022 minutes, it will complete 252 full eight-minute cycles and have 6 minutes left. Sinc |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second digit does not have to be different. For example, 101 and 222 have this form, while 220 and 123 do not. Similarly, we define the form of the number $\overline{\overline{b a b c}}$. How many numbers of the form $\overline{a b a b c}$ are divisible by 5? | Numbers divisible by $5$ are those ending in 0 or 5, so we have two options for $c$. For $a$, we have 9 options, as the number cannot start with zero, and the value of $b$ can be anything.
Thus, we get that the total number of such numbers is $2 \cdot 9 \cdot 10=180$. | 180 | Number Theory | olympiads | null | null | \(\boxed{180}\) | 3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second digit does not have to be different. For example, 101 and 222 have this form, while 220 and 123 do not. Similarly, we define the form of the number $\overline{\overline{b a b c}}$. How many numbers of the form $\overline{a b a b c}$ are divisible by 5?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Numbers divisible by $5$ are those ending in 0 or 5, so we have two options for $c$. For $a$, we have 9 options, as the number cannot |
4. What is the minimum number of chips that can be placed in the cells of a $99 \times 99$ table so that in each $4 \times 4$ square there are at least eight chips? | Add a row and a column with number 100 to the table. Place a chip in each of their cells. Divide the expanded table into 625 squares $4 \times 4$. In each square, there can be no more than eight empty cells, so there are no more than 5000 empty cells in the entire table. Therefore, the total number of chips must be at least $99^{2}-5000=4801$.
Now, place chips in the cells of the expanded table where the product of the coordinates is divisible by four. In this case, each of the 625 squares $4 \times 4$ will have exactly 8 chips. Moreover, the added row and column will be completely filled with chips. Therefore, in the original table, there will be $625 \cdot 8-199=4801$ chips. | 4801 | Combinatorics | olympiads | null | null | \(\boxed{4801}\) | 4. What is the minimum number of chips that can be placed in the cells of a $99 \times 99$ table so that in each $4 \times 4$ square there are at least eight chips?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Add a row and a column with number 100 to the table. Place a chip in each of their cells. Divide the expanded table into 625 squares $4 \times 4$. In each square, there can be no more than eight empty cells, so there are no more than 5000 empty cells in the entire table. Therefore, the total number of chips must be at least $99^{2}-5000=4801 |
1. (10 points) We will call a date diverse if in its representation in the form DD/MM/YY (day-month-year) all digits from 0 to 5 are present. How many diverse dates are there in the year 2013? | Note that in any date of 2013 in the specified format, the digits 1 and 3 are used, so for the day and month of a diverse date, the digits left are 0, 2, 4, and 5.
Let $Д_{1}$ and $Д_{2}$ be the first and second digits in the day's notation, and $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ be the first and second digits in the month's notation. Since there are 12 months in a year, $\mathrm{M}_{1} \leqslant 1$, hence $\mathrm{M}_{1}=0$. Moreover, no month has more than 31 days, and of the remaining digits, only 2 is suitable for $Д_{1}$. The digits 4 and 5 can be placed in positions $Д_{2}$ and $\mathrm{M}_{2}$ in any order, giving a valid date. | 2 | Combinatorics | olympiads | null | null | \(\boxed{2}\) | 1. (10 points) We will call a date diverse if in its representation in the form DD/MM/YY (day-month-year) all digits from 0 to 5 are present. How many diverse dates are there in the year 2013?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Note that in any date of 2013 in the specified format, the digits 1 and 3 are used, so for the day and month of a diverse date, the digits left are 0, 2, 4, and 5.
Let $Д_{1}$ and $Д_{2}$ be the first and second digits in the day's notation, and $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ be the first and second digits in the |
2. (15 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different four-letter words can be formed from the letters of the word КАША? And from the letters of the word ХЛЕБ? In your answer, indicate the sum of the found numbers. | In the word ХЛЕБ, all letters are different. Therefore, by rearranging the letters, we get $4 \cdot 3 \cdot 2 \cdot 1=24$ words. From the word КАША, we can form 12 words. Indeed, for the letters K and Ш, there are $4 \cdot 3=12$ positions, and we write the letters А in the remaining places. Thus, in total, we get $24+12=36$ words. | 36 | Combinatorics | olympiads | null | null | \(\boxed{36}\) | 2. (15 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different four-letter words can be formed from the letters of the word КАША? And from the letters of the word ХЛЕБ? In your answer, indicate the sum of the found numbers.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
In the word ХЛЕБ, all letters are different. Therefore, by rearranging the letters, we get $4 \cdot 3 \cdot 2 \cdot 1=24$ words. From the word КАША, we can form 12 wo |
7. (50 points) From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property? | Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this group. Suppose there is another number $C$, and $j=C \bmod 5$. Then the numbers $i+j$ and $5-i+j$ are divisible by 5, as well as their difference, which is $2 i-5$, which is impossible for $i \neq 0$.
Now consider the alternative case, where all numbers in the group are divisible by 5. In the given range, there are exactly 40 numbers that are multiples of 5, and all these numbers can be included in the group simultaneously. | 40 | Number Theory | olympiads | null | null | \(\boxed{40}\) | 7. (50 points) From the numbers 1 to 200, one or several were selected into a separate group with the following property: if there are at least two numbers in the group, then the sum of any two numbers in this group is divisible by 5. What is the maximum number of numbers that can be in a group with this property?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Suppose that the group selected a number $A$, which gives a remainder $i \neq 0$ when divided by 5. If there is another number $B$ in the group, then it must give a remainder $5-i$ when divided by 5, so that $A+B$ is divisible by 5. We will show that there cannot be any other numbers in this group. Suppose there is another number $C$, and $j=C \bmod 5$. Then |
8. (50 points) Each of the 12 knights sitting around a round table has thought of a number, and all the numbers are different. Each knight claims that the number they thought of is greater than the numbers thought of by their neighbors to the right and left. What is the maximum number of these claims that can be true? | Let's renumber the knights in order with numbers from 1 to 12. In the pairs $(1,2),(3,4)$, $\ldots,(11,12)$, at least one of the knights is lying (specifically, the one who guessed the smaller number). Therefore, there can be no more than 6 true statements. Now let's provide an example where exactly 6 statements are true. Suppose the knights with odd numbers guessed the numbers from 1 to 6, and those with even numbers guessed the numbers from 7 to 12 (in any order). Then all the knights with even numbers are telling the truth. | 6 | Logic and Puzzles | olympiads | null | null | \(\boxed{6}\) | 8. (50 points) Each of the 12 knights sitting around a round table has thought of a number, and all the numbers are different. Each knight claims that the number they thought of is greater than the numbers thought of by their neighbors to the right and left. What is the maximum number of these claims that can be true?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's renumber the knights in order with numbers from 1 to 12. In the pairs $(1,2),(3,4)$, $\ldots,(11,12)$, at least one of the knights is lying (specifically, the one who guessed the smaller number). Therefore, there can be no more than 6 true statements. Now let' |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a+b}{c}\right)^{4}+\left(\frac{b+c}{d}\right)^{4}+\left(\frac{c+d}{a}\right)^{4}+\left(\frac{d+a}{b}\right)^{4}
$$ | By Cauchy's inequality for means,
$$
A \geqslant 4 \cdot \frac{(a+b)(b+c)(c+d)(d+a)}{a b c d}=64 \cdot \frac{a+b}{2 \sqrt{a b}} \cdot \frac{b+c}{2 \sqrt{b c}} \cdot \frac{c+d}{2 \sqrt{c d}} \cdot \frac{d+a}{2 \sqrt{d a}} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$. | 64 | Inequalities | olympiads | null | null | \(\boxed{64}\) | 2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a+b}{c}\right)^{4}+\left(\frac{b+c}{d}\right)^{4}+\left(\frac{c+d}{a}\right)^{4}+\left(\frac{d+a}{b}\right)^{4}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By Cauchy's inequality for means,
$$
A \geqslant 4 \cdot \frac{(a+b)(b+c)(c+d)(d+a)}{a b c d}=64 \cdot \frac{a+b}{2 \sqrt{a b}} \cdot \fr |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a^{2}+b^{2}}{c d}\right)^{4}+\left(\frac{b^{2}+c^{2}}{a d}\right)^{4}+\left(\frac{c^{2}+d^{2}}{a b}\right)^{4}+\left(\frac{d^{2}+a^{2}}{b c}\right)^{4}
$$ | By the Cauchy inequalities for means,
$$
A \geqslant 4 \cdot \frac{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+d^{2}\right)\left(d^{2}+a^{2}\right)}{c d \cdot a d \cdot a b \cdot b c}=64 \cdot \frac{a^{2}+b^{2}}{2 a b} \cdot \frac{b^{2}+c^{2}}{2 b c} \cdot \frac{c^{2}+d^{2}}{2 c d} \cdot \frac{d^{2}+a^{2}}{2 d a} \geqslant 64
$$
Equality is achieved when $a=b=c=d=1$. | 64 | Inequalities | olympiads | null | null | \(\boxed{64}\) | 2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(\frac{a^{2}+b^{2}}{c d}\right)^{4}+\left(\frac{b^{2}+c^{2}}{a d}\right)^{4}+\left(\frac{c^{2}+d^{2}}{a b}\right)^{4}+\left(\frac{d^{2}+a^{2}}{b c}\right)^{4}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
By the Cauchy inequalities for means,
$$
A \geqslant 4 \cdot \frac{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+d^{2}\right)\left(d^{2}+a^{2}\right)}{c d \cdot a d \cdot a b \cdo |
1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, as each of them is counted twice - in the row and in the column. Then
$$
S \geqslant \frac{1}{2}(5+6+\ldots+13+14)=\frac{1}{2} \cdot 95=47 \frac{1}{2}, \quad \text { that is } \quad S \geqslant 48 .
$$
An example for $S=48$ is given below.
| 1 | 1 | 1 | 1 | 1 |
| :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 1 | 2 | 2 |
| 1 | 1 | 2 | 3 | 3 |
| 1 | 2 | 2 | 3 | 3 |
| 2 | 3 | 3 | 3 | 4 | | 48 | Combinatorics | olympiads | null | null | \(\boxed{48}\) | 1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, as each of them is counte |
1. In the cells of a $4 \times 6$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 4. Since all these sums are different, the minimal possible set of their values is $\{4,5, \ldots, 12,13\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, since each of them is counted twice - in the row and in the column. Then
$$
S \geqslant \frac{1}{2}(4+5+\ldots+12+13)=\frac{1}{2} \cdot 85=42 \frac{1}{2}, \quad \text { that is } \quad S \geqslant 43
$$
An example for $S=43$ is given below.
| 1 | 1 | 1 | 1 | 1 | 2 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 1 | 2 | 2 | 3 |
| 1 | 1 | 2 | 3 | 3 | 2 |
| 1 | 2 | 2 | 2 | 3 | 4 | | 43 | Combinatorics | olympiads | null | null | \(\boxed{43}\) | 1. In the cells of a $4 \times 6$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 4. Since all these sums are different, the minimal possible set of their values is $\{4,5, \ldots, 12,13\}$. By adding the sums of the rows and columns of the table, we get twice the sum $S$ of all the numbers in the table, since each of them is coun |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a^{2}+\frac{1}{b c}\right)^{3}+\left(b^{2}+\frac{1}{c d}\right)^{3}+\left(c^{2}+\frac{1}{d a}\right)^{3}+\left(d^{2}+\frac{1}{a b}\right)^{3}
$$ | We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(\frac{2 a}{\sqrt{b c}}\right)^{3}+\left(\frac{2 b}{\sqrt{c d}}\right)^{3}+\left(\frac{2 c}{\sqrt{d a}}\right)^{3}+\left(\frac{2 d}{\sqrt{a b}}\right)^{3} \geqslant 32\left(\frac{a^{2}}{b c} \cdot \frac{b^{2}}{c d} \cdot \frac{c^{2}}{d a} \cdot \frac{d^{2}}{a b}\right)^{3 / 8}=32
$$
Equality is achieved when $a=b=c=d=1$. | 32 | Algebra | olympiads | null | null | \(\boxed{32}\) | 2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a^{2}+\frac{1}{b c}\right)^{3}+\left(b^{2}+\frac{1}{c d}\right)^{3}+\left(c^{2}+\frac{1}{d a}\right)^{3}+\left(d^{2}+\frac{1}{a b}\right)^{3}
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(\frac{2 a}{\sqrt{b c}}\right)^{3}+\left(\frac{2 b}{\sqrt{c d}}\right)^{3}+\left(\frac{2 c}{\sqrt{ |
4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible. | Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that
$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a_{7}+a_{k} a_{8} \quad \text { for any } \quad k \in\{0, \ldots, 8\}
$$
Clearly, $b_{0}=a_{0} a_{8} \bmod 10$. Since $10^{17}>m n \geqslant a_{0} a_{8} \cdot 10^{16}$, we get $a_{0} a_{8} \leqslant 9$, which means $b_{0}=a_{0} a_{8}$. Suppose that for some $k0$, we get $a_{0}=1$ and $a_{k}=0$ for $k=1, \ldots, 6$. The number $m=220000001$ clearly satisfies the condition of the problem. | 220000001 | Number Theory | olympiads | null | null | \(\boxed{220000001}\) | 4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that
$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a_{7}+a_{k} a_{8} \quad \text { for any } \quad k \ |
1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There turned out to be $B$ such numbers. What is $100 B+A$? | We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 to 9, giving us 9 numbers. And if $y=5$, then $x$ can take values only from 6 to 9, giving us another 4 numbers. In total, we get 13 numbers.
Let's calculate $B$. Here, there can only be the case where $y=5$, because $x$ cannot be less than 0. So, $x$ can take values from 1 to 4. Therefore, $B=4$. From this, we get that $100 B+A=400+13=413$. | 413 | Number Theory | olympiads | null | null | \(\boxed{413}\) | 1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There turned out to be $B$ such numbers. What is $100 B+A$?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 to 9, giving us 9 number |
2. (20 points) Find the number of different four-digit numbers that can be obtained by rearranging the digits of the number 2021 (including this number itself). | The number of options can be found by enumerating the permutations of the digits: 2021, 2012, 2201, 2210, 2102, 2120, 1022, 1202, 1220.
The number of options can also be calculated using combinatorial methods. The position of zero can be chosen in three ways, as it should not be the first. Then, in three ways, we choose the position of 1, and on the two remaining places, we put the twos. In the end, we get $3 \cdot 3=9$ possible numbers. | 9 | Combinatorics | olympiads | null | null | \(\boxed{9}\) | 2. (20 points) Find the number of different four-digit numbers that can be obtained by rearranging the digits of the number 2021 (including this number itself).
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The number of options can be found by enumerating the permutations of the digits: 2021, 2012, 2201, 2210, 2102, 2120, 1022, 1202, 1220.
The number of options can also be calculated using combinatorial methods. The positi |
3. (20 points) At the Journalism Faculty of the University of Enchanted Commonwealth, 4 chickens are applying. The faculty has 2 places in the daytime program and 3 places in the evening program. Assuming all 4 chickens will be admitted to the faculty, determine the number of outcomes in which exactly two chickens will be admitted to the evening program. | The faculty has a total of 5 places, to which 4 applicants are applying. Since exactly two chickens will be admitted to the evening department, the other two will be admitted to the daytime department. The number of ways to choose 2 out of 4 applicants to be admitted to the daytime department is $C_{4}^{2}=6$. | 6 | Combinatorics | olympiads | null | null | \(\boxed{6}\) | 3. (20 points) At the Journalism Faculty of the University of Enchanted Commonwealth, 4 chickens are applying. The faculty has 2 places in the daytime program and 3 places in the evening program. Assuming all 4 chickens will be admitted to the faculty, determine the number of outcomes in which exactly two chickens will be admitted to the evening program.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The faculty has a total of 5 places, to which 4 applicants are applying. Since exactly two chickens will be admitted to the evening department, the other tw |
9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take? | Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then
$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$
Therefore,
$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$
Similarly for the second and third terms, we get
$$
\frac{b\left(c^{2}+3\right)}{b+c}=b(c+a), \quad \frac{c\left(a^{2}+3\right)}{c+a}=c(a+b)
$$
Thus,
$$
\begin{gathered}
\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}=a(b+c)+b(c+a)+c(a+b)= \\
\quad=a b+a c+b c+b a+c a+c b=2(a b+b c+c a)=2 \times 3=6
\end{gathered}
$$ | 6 | Algebra | olympiads | null | null | \(\boxed{6}\) | 9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then
$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$
Therefore,
$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$
Similarly for the second and third terms, we get
$$
\frac{b\left(c^ |
1. Given the quadratic trinomial $f(x)=a x^{2}-a x+1$. It is known that $|f(x)| \leqslant 1$ for all $x \in[0,1]$. What is the greatest value that $a$ can take? | It is not difficult to check that $a=8$ works. Indeed, $|2 x-1| \leqslant 1$ for $x \in[0,1]$, so $f(x)=8 x^{2}-8 x+1=2(2 x-1)^{2}-1 \leqslant 1$, and the inequality $f(x) \geqslant-1$ holds for all $x$.
Suppose that $a>8$. Then
$$
f\left(\frac{1}{2}\right)=\frac{a}{4}-\frac{a}{2}+1=1-\frac{a}{4}=\frac{4-a}{4}<-1
$$
which is impossible by the condition. | 8 | Algebra | olympiads | null | null | \(\boxed{8}\) | 1. Given the quadratic trinomial $f(x)=a x^{2}-a x+1$. It is known that $|f(x)| \leqslant 1$ for all $x \in[0,1]$. What is the greatest value that $a$ can take?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
It is not difficult to check that $a=8$ works. Indeed, $|2 x-1| \leqslant 1$ for $x \in[0,1]$, so $f(x)=8 x^{2}-8 x+1=2(2 x-1)^{2}-1 \leqslant 1$, and the inequality $f(x) \geqsla |
5. (20 points) At a market in Egypt, a tourist is bargaining with a seller over a souvenir worth 10000 Egyptian pounds. The tourist first reduces the price by x percent $(0<x<100)$, then the seller increases the price by $x$ percent, and so on. The number $x$ remains constant throughout the bargaining, and the seller increases the price at least once. The bargaining continues until one of the participants receives a non-integer value for the price of the souvenir. Find the maximum possible number of price changes during such a bargaining session (including the final non-integer price). | The final cost of the souvenir can be found using one of two formulas (depending on who had the last word): $10000 \cdot\left(1-\frac{x}{100}\right)^{n} \cdot\left(1+\frac{x}{100}\right)^{n}$ or $10000 \cdot\left(1-\frac{x}{100}\right)^{n+1} \cdot\left(1+\frac{x}{100}\right)^{n}$.
After some transformations, we get $\frac{(100-x)^{n}(100+x)^{n}}{100^{2 n-2}}$ and $\frac{(100-x)^{n+1}(100+x)^{n}}{100^{2 n-2}}$.
An expression of the form $(100-x)^{a}(100+x)^{b}$ is divisible by $100^{t}$ for $a>0, b>0, x>0$ and $t>0$ at least once for $x=10 i, i=1, \ldots, 9$ (an even number, a multiple of 5). | 5 | Algebra | olympiads | null | null | \(\boxed{5}\) | 5. (20 points) At a market in Egypt, a tourist is bargaining with a seller over a souvenir worth 10000 Egyptian pounds. The tourist first reduces the price by x percent $(0<x<100)$, then the seller increases the price by $x$ percent, and so on. The number $x$ remains constant throughout the bargaining, and the seller increases the price at least once. The bargaining continues until one of the participants receives a non-integer value for the price of the souvenir. Find the maximum possible number of price changes during such a bargaining session (including the final non-integer price).
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The final cost of the souvenir can be found using one of two formulas (depending on who had the last word): $10000 \cdot\left(1-\frac{x}{100}\right)^{n} \cdot\left(1+\frac{x}{100}\right)^{n}$ or $10000 \cdot\left(1-\frac{x}{100}\right)^{n+1} \cdot\left(1+\frac{x}{100}\right)^{n}$.
After some transf |
9. (40 points) What is the maximum number of numbers that can be chosen among the natural numbers from 1 to 3000 such that the difference between any two of them is different from 1, 4, and 5? | Let's provide an example. We can choose all numbers divisible by 3. Then the difference between any two numbers will also be divisible by 3, while the numbers 1, 4, and 5 are not divisible by 3.
The estimate is based on the consideration that among 6 consecutive numbers, 3 numbers cannot be chosen. Let's prove this statement. Take any chosen number and the five numbers following it. Then the second, fifth, and sixth numbers are definitely not chosen. The third and fourth numbers remain, but they cannot both be chosen at the same time. It is even more impossible to choose 3 numbers among five or fewer consecutive numbers. | 1000 | Combinatorics | olympiads | null | null | \(\boxed{1000}\) | 9. (40 points) What is the maximum number of numbers that can be chosen among the natural numbers from 1 to 3000 such that the difference between any two of them is different from 1, 4, and 5?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's provide an example. We can choose all numbers divisible by 3. Then the difference between any two numbers will also be divisible by 3, while the numbers 1, 4, and 5 are not divisible by 3.
The estimate is based on the consideration that among 6 consecutive numbers, 3 numbers cannot be chosen. Let's prove t |
4. Find the value of the expression
$$
\frac{1}{1+m+m n}+\frac{1}{1+n+n k}+\frac{1}{1+k+k m} \text { given that } m=\frac{1}{n k} \text {. }
$$ | ## Solution.
$$
\begin{gathered}
\frac{1}{1+m+m n}+\frac{1}{1+n+n k}+\frac{1}{1+k+k m}=\frac{k}{k+k m+k m n}+\frac{k m}{k m+k m n+k m n k}+\frac{1}{1+k+k m}= \\
=\frac{k}{k+k m+1}+\frac{k m}{k m+1+k}+\frac{1}{1+k+k m}=\frac{k+k m+1}{1+k+k m}=1
\end{gathered}
$$ | 1 | Algebra | olympiads | null | null | \(\boxed{1}\) | 4. Find the value of the expression
$$
\frac{1}{1+m+m n}+\frac{1}{1+n+n k}+\frac{1}{1+k+k m} \text { given that } m=\frac{1}{n k} \text {. }
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Solution.
$$
\begin{gathered}
\frac{1}{1+m+m n}+\frac{1}{1+n+n k}+\frac{1}{1+k+k m}=\frac{k}{k+k m+k m n}+\frac{k m}{k m+k m n+ |
3. Find $g$(2021), if for any real $x, y$ the equality holds
$$
g(x-y)=g(x)+g(y)-2022(x+y)
$$ | Substitute $x=y=0$, we get
$$
g(0)=g(0)+g(0)-2022(0+0) \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=g(x)+g(x)-2022(x+x) \Rightarrow g(x)=2022 x \Rightarrow \\
g(2021)=2022 \cdot 2021=4086462 .
\end{gathered}
$$ | 4086462 | Algebra | olympiads | null | null | \(\boxed{4086462}\) | 3. Find $g$(2021), if for any real $x, y$ the equality holds
$$
g(x-y)=g(x)+g(y)-2022(x+y)
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Substitute $x=y=0$, we get
$$
g(0)=g(0)+g(0)-2022(0+0) \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathe |
3. For the quadratic trinomials $f_{1}(x)=a x^{2}+b x+c_{1}, f_{2}(x)=a x^{2}+b x+c_{2}$, $\ldots, f_{2020}(x)=a x^{2}+b x+c_{2020}$, it is known that each of them has two roots.
Denote by $x_{i}$ one of the roots of $f_{i}(x)$, where $i=1,2, \ldots, 2020$. Find the value
$$
f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\cdots+f_{2020}\left(x_{2019}\right)+f_{1}\left(x_{2020}\right)
$$ | ## Solution:
Since $f_{1}\left(x_{1}\right)=0$, then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$.
Similarly, we can obtain the following equalities:
$$
f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_{2020}\left(x_{2019}\right)=c_{2020}-c_{2019}, f_{1}\left(x_{2020}\right)=c_{1}-c_{2020}
$$
Adding these equalities, we get
$$
f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\cdots+f_{2020}\left(x_{2019}\right)+f_{1}\left(x_{2020}\right)=0
$$ | 0 | Algebra | olympiads | null | null | \(\boxed{0}\) | 3. For the quadratic trinomials $f_{1}(x)=a x^{2}+b x+c_{1}, f_{2}(x)=a x^{2}+b x+c_{2}$, $\ldots, f_{2020}(x)=a x^{2}+b x+c_{2020}$, it is known that each of them has two roots.
Denote by $x_{i}$ one of the roots of $f_{i}(x)$, where $i=1,2, \ldots, 2020$. Find the value
$$
f_{2}\left(x_{1}\right)+f_{3}\left(x_{2}\right)+\cdots+f_{2020}\left(x_{2019}\right)+f_{1}\left(x_{2020}\right)
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Solution:
Since $f_{1}\left(x_{1}\right)=0$, then $f_{2}\left(x_{1}\right)=f_{1}\left(x_{1}\right)+\left(c_{2}-c_{1}\right)=c_{2}-c_{1}$.
Similarly, we can obtain the following equalities:
$$
f_{3}\left(x_{2}\right)=c_{3}-c_{2}, \ldots, f_ |
3. Find $g(2021)$, if for any real $x, y$ the equality holds
$$
g(x-y)=2021(g(x)+g(y))-2022 x y
$$ | Substitute $x=y=0$, we get
$$
g(0)=2021(g(0)+g(0))-2022 \cdot 0 \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=2021(g(x)+g(x))-2022 \cdot x^{2} \Rightarrow g(x)=\frac{2022 x^{2}}{2 \cdot 2021}=\frac{1011 x^{2}}{2021} \Rightarrow \\
g(2021)=\frac{1011 \cdot 2021^{2}}{2021}=1011 \cdot 2021=2043231
\end{gathered}
$$ | 2043231 | Algebra | olympiads | null | null | \(\boxed{2043231}\) | 3. Find $g(2021)$, if for any real $x, y$ the equality holds
$$
g(x-y)=2021(g(x)+g(y))-2022 x y
$$
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Substitute $x=y=0$, we get
$$
g(0)=2021(g(0)+g(0))-2022 \cdot 0 \Rightarrow g(0)=0
$$
Substitute $x=y$, we get
$$
\begin{gathered}
g(0)=2021(g(x)+g(x))-2022 \cdot x^{2} |
2. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2018}{2019}\right)$.
(7 points). | ## Solution:
Substitute $\frac{1}{x}$ for $x$ in the original equation. Together with the original equation, we get a system of linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$.
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1} \\
\left(\frac{1}{x}-1\right) f\left(\frac{1}{x}\right)+f(x)=\frac{x}{x-1}
\end{array}\right.
$$
Solving the obtained system, we find $f(x)=\frac{1}{1-x}$. Therefore, $f\left(\frac{2018}{2019}\right)=\frac{1}{1-\frac{2018}{2019}}=2019$. | 2019 | Algebra | olympiads | null | null | \(\boxed{2019}\) | 2. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2018}{2019}\right)$.
(7 points).
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Solution:
Substitute $\frac{1}{x}$ for $x$ in the original equation. Together with the original equation, we get a system of linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$.
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\r |
3. For any pair of numbers, a certain operation «*» is defined, satisfying the following properties: $a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation «$\cdot$» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 2=2018$. | ## Solution:
Given the condition of the problem, we have $x * 1=x *(x * x)=(x * x) \cdot x=1 \cdot x=x$. Then
1) $(x * 2) \cdot 2=2018 \cdot 2=4036$,
2) $(x * 2) \cdot 2=x *(2 * 2)=x \cdot 1=x$.
Therefore, $x=4036$. | 4036 | Algebra | olympiads | null | null | \(\boxed{4036}\) | 3. For any pair of numbers, a certain operation «*» is defined, satisfying the following properties: $a *(b * c)=(a * b) \cdot c$ and $a * a=1$, where the operation «$\cdot$» is the multiplication operation. Find the root $x$ of the equation: $\quad x * 2=2018$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Solution:
Given the condition of the problem, we have $x * 1=x *(x * x)=(x * x) \cdot x=1 \cdot x=x$. The |
3. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, it is known that
$$
f\left(\frac{a-b-c}{2 a}\right)=f\left(\frac{c-a-b}{2 a}\right)=0
$$
Find the value of the product $f(-1) \cdot f(1)$. | ## Solution:
$$
f\left(\frac{a-b-c}{2 a}\right)=\frac{a(a-b-c)^{2}}{4 a^{2}}+\frac{b(a-b-c)}{2 a}+c=\frac{(a-b+c)(a+b+c)}{4 a}=\frac{f(-1) \cdot f(1)}{4 a}=0
$$ | 0 | Algebra | olympiads | null | null | \(\boxed{0}\) | 3. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, it is known that
$$
f\left(\frac{a-b-c}{2 a}\right)=f\left(\frac{c-a-b}{2 a}\right)=0
$$
Find the value of the product $f(-1) \cdot f(1)$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Solution:
$$
f\left(\frac{a-b-c}{2 a}\right)=\frac{a(a-b-c)^{2}}{4 a^{2}}+\f |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions:
$a^{3}-2022 a+1011=0, \quad b^{3}-2022 b+1011=0, \quad c^{3}-2022 c+1011=0$. | ## Solution.
The cubic equation $t^{3}-2022 t+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t+1011: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2022 \\
a b c=-1011
\end{array}\right.
$$
We find the value of the expression: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+a c}{a b c}=\frac{-2022}{-1011}=2$. | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions:
$a^{3}-2022 a+1011=0, \quad b^{3}-2022 b+1011=0, \quad c^{3}-2022 c+1011=0$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
## Solution.
The cubic equation $t^{3}-2022 t+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t+1011: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\le |
2. For the quadratic trinomial $p(x)=(a-1) x^{2}-(a-1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$. | Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the minimum value of $p(x)$ is $p\left(\frac{1}{2}\right)=2022-\frac{(a-1)}{4}$. The maximum possible value of $a$ will be achieved when $p\left(\frac{1}{2}\right)=-2022$. Therefore,
$$
2022-\frac{(a-1)}{4}=-2022 \Rightarrow \frac{(a-1)}{4}=4044 \Rightarrow a-1=16176 \Rightarrow a=16177
$$ | 16177 | Algebra | olympiads | null | null | \(\boxed{16177}\) | 2. For the quadratic trinomial $p(x)=(a-1) x^{2}-(a-1) x+2022$, it is known that $-2022 \leq p(x) \leq 2022$ for $x \in[0 ; 1]$. Find the greatest possible value of $a$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Since $p(0)=p(1)=2022$, the graph of the quadratic trinomial is a parabola symmetric about the line $x=\frac{1}{2}$. From the conditions that $-2022 \leq$ $p(x) \leq 2022$ for $x \in[0 ; 1]$ and $p(0)=p(1)=2022$, it follows that the branches of the parabola are directed upwards. Then the |
3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three distinct real numbers satisfying the conditions:
$a^{3}-2020 a+1010=0, \quad b^{3}-2020 b+1010=0, c^{3}-2020 c+1010=0$. | The cubic equation $t^{3}-2020 t+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t+1010: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2020 \\
a b c=-1010
\end{array}\right.
$$
We find the value of the expression: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+a c}{a b c}=\frac{-2020}{-1010}=2$. | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three distinct real numbers satisfying the conditions:
$a^{3}-2020 a+1010=0, \quad b^{3}-2020 b+1010=0, c^{3}-2020 c+1010=0$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
The cubic equation $t^{3}-2020 t+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t+1010: f(-100)0, f(10)0\right)$.
Let these roots be $a, b, c$. Then, by Vieta's theorem:
$$
\left\{\b |
4. (4 points) Oleg usually arrives on a business trip by the 11 o'clock train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine the time Oleg met the car. | hours 55 minutes.
$+(4$ points) - solution is correct (by any method)
+- (3 points) - solution is correct, but there are arithmetic errors
-+ (2 points) - there are reasonable ideas in solving the problem, but the problem is not solved in general | 10 | Logic and Puzzles | olympiads | null | null | \(\boxed{10}\) | 4. (4 points) Oleg usually arrives on a business trip by the 11 o'clock train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine the time Oleg met the car.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
hours 55 minutes.
$+(4$ points) - solution is correct (by any method)
+- (3 points) - solution is correct, but there are a |
5. (4 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$,
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \ldots, 2008$.
Calculate the value of $x_{2009} y_{2008}+x_{2008} y_{2009}$. | .
$+(4$ points) - solution is correct
+- (3 points) - the idea of the recurrence relation is proven, but there are arithmetic errors in the solution
-+ (2 points) - the idea of the recurrence relation is formulated, but the problem is not completed | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 5. (4 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$,
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \ldots, 2008$.
Calculate the value of $x_{2009} y_{2008}+x_{2008} y_{2009}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
.
$+(4$ points) - solution is correct
+- (3 points) - the idea of the recurrence relation is proven, but there are arithmeti |
2. (3 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \ldots, 2008$.
Calculate the value of $x_{2009} y_{2008}+x_{2008} y_{2009}$. | .
$+(3$ points) - the solution is correct
+- (2 points) - the idea of the recurrence relation is proven, but there are arithmetic errors in the solution
-+ (1 point) - the idea of the recurrence relation is formulated, but the problem is not completed | 2 | Algebra | olympiads | null | null | \(\boxed{2}\) | 2. (3 points) Let two sequences of numbers $\left(x_{0}, x_{1}, \ldots, x_{2009}\right)$, $\left(y_{0}, y_{1}, \ldots, y_{2009}\right)$ be constructed according to the following rules:
a) $x_{0}=12, x_{1}=\frac{1}{3}, y_{0}=4, y_{1}=\frac{1}{18}$
b) $x_{i+1}=x_{i-1}+4 x_{i}$ and $y_{i+1}=y_{i-1}-4 y_{i}$ for $i=1, \ldots, 2008$.
Calculate the value of $x_{2009} y_{2008}+x_{2008} y_{2009}$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
.
$+(3$ points) - the solution is correct
+- (2 points) - the idea of the recurrence relation is proven, but there are arithm |
1. (3 points) Oleg usually arrives on a business trip by the 11 AM train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine the time Oleg met the car. | hours 55 minutes.
+ (3 points) - the solution is correct (by any method)
$+-(2$ points) - the solution is correct, but there are arithmetic errors
-+ (1 point) - there are reasonable ideas in solving the problem, but the problem is not solved overall | 10 | Logic and Puzzles | olympiads | null | null | \(\boxed{10}\) | 1. (3 points) Oleg usually arrives on a business trip by the 11 AM train. A car arrives at the station for this train. This time, the train arrived an hour earlier, and Oleg started walking towards the car. Meeting the car on the way, he got in, and as a result, arrived 10 minutes earlier than planned. Determine the time Oleg met the car.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
hours 55 minutes.
+ (3 points) - the solution is correct (by any method)
$+-(2$ points) - the solution is correct, but there |
15. Hydrogen was passed over a heated powder (X1). The resulting red substance (X2) was dissolved in concentrated sulfuric acid. The resulting solution of the substance blue (X3) was neutralized with potassium hydroxide - a blue precipitate (X4) formed, which upon heating turned into a black powder (X1). What substances are involved in the described process? Indicate the molar mass of the initial and final substance (X1). | g/mol; $\mathrm{m}(\mathrm{CuO})=80$ g/mol; $\mathrm{X}_{1}-\mathrm{CuO} ; \mathrm{X}_{2}-\mathrm{Cu} ; \mathrm{X}_{3}-\mathrm{CuSO}_{4 ;} \mathrm{X}_{4}-\mathrm{Cu}(\mathrm{OH})_{2}$ | 80 | Other | olympiads | null | null | \(\boxed{80}\) | 15. Hydrogen was passed over a heated powder (X1). The resulting red substance (X2) was dissolved in concentrated sulfuric acid. The resulting solution of the substance blue (X3) was neutralized with potassium hydroxide - a blue precipitate (X4) formed, which upon heating turned into a black powder (X1). What substances are involved in the described process? Indicate the molar mass of the initial and final substance (X1).
The following text is the beginning part of the answer, which you can refer to for solving the problem:
g/mol; $\mathrm{m}(\mathrm{CuO})=80$ g/mol; $\mathrm{X}_{1}-\mathrm{CuO} ; \mathrm{X}_{2}-\m |
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum number of rounds after which an early winner can appear?
## Answers and solutions:
## Translation of the question and answers into English:
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum number of rounds after which an early winner can appear?
## Answers and solutions: | Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one participant with more than three points. Since there are still 3 rounds ahead, the winner is still unknown.
Example. Suppose in the first 7 rounds, the leader won all their matches, and all other matches ended in draws. Then two students who have not yet faced the leader have 3.5 points each, while the others have 3 points each. Since only 2 rounds remain until the end of the tournament, the winner is already determined. | 7 | Combinatorics | olympiads | null | null | \(\boxed{7}\) | 16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum number of rounds after which an early winner can appear?
## Answers and solutions:
## Translation of the question and answers into English:
16. At the quiz in the Museum of Entertaining Sciences of SFU, 10 schoolchildren are participating. In each round, the students are divided into pairs. Each participant meets every other participant exactly once. A win in a match earns 1 point, a draw earns 0.5 points, and a loss earns 0 points. What is the minimum number of rounds after which an early winner can appear?
## Answers and solutions:
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one participant with more than three points. Since there are still 3 rounds ahead, the winner is sti |
16. A worker at an aluminum plant can produce 16 blanks or 10 parts from blanks in one shift. It is known that exactly one part is made from each blank. What is the maximum number of blanks the worker can produce in one shift to make parts from them in the same shift? | Let's denote by $x$ the duration of the worker's working day in hours. Then the worker makes one part from a blank in $\frac{x}{10}$ hours, one blank in $\frac{x}{16}$ hours, and a blank and a part from it in $\frac{x}{10}+\frac{x}{16}=\frac{13 x}{80}$ hours. Since $x: \frac{13 x}{80}=6 \frac{2}{13}$, the maximum number of blanks that the worker can make in a day so that parts can be made from all of them on the same day is 6. | 6 | Logic and Puzzles | olympiads | null | null | \(\boxed{6}\) | 16. A worker at an aluminum plant can produce 16 blanks or 10 parts from blanks in one shift. It is known that exactly one part is made from each blank. What is the maximum number of blanks the worker can produce in one shift to make parts from them in the same shift?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let's denote by $x$ the duration of the worker's working day in hours. Then the worker makes one part from a blank in $\frac{x}{10}$ hours, one blank in $\frac{x}{16}$ hours, and a blank and a part from it in $\frac |
16. Every sixth bus in the bus park of the aluminum plant is equipped with an air conditioner. After the plant director ordered to install it on 5 more buses, a quarter of the buses had an air conditioner. How many buses are there in the park of the plant, if each bus is equipped with only one air conditioner? | Let the number of buses in the park be denoted by $x$. Then, $\frac{1}{6} x + 5$ buses are equipped with air conditioning. This constitutes a quarter of all the buses in the park. We have the equation: $\left(\frac{1}{6} x + 5\right) \cdot 4 = x$ or $\frac{1}{3} x = 20$. From this, $x = 60$. | 60 | Algebra | olympiads | null | null | \(\boxed{60}\) | 16. Every sixth bus in the bus park of the aluminum plant is equipped with an air conditioner. After the plant director ordered to install it on 5 more buses, a quarter of the buses had an air conditioner. How many buses are there in the park of the plant, if each bus is equipped with only one air conditioner?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let the number of buses in the park be denoted by $x$. Then, $\frac{1}{6} x + 5$ buses are equipped with air conditioning. This constitutes a quar |
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$. | Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{d c b a}=1000 d+100 c+10 b+a$, where $a$, $b, c, d$ are digits and $a \neq 0$.
According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.
We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from which, since $a$ and $d$ are digits and $a \neq 0, 1 \leq a+d \leq 18$, thus $a+d=10$. Further, $1001 \cdot 10+110(b+c) \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19$, thus, $b+c=9$.
Therefore, $N=X+Y=1001 \cdot 10+110 \cdot 9=11000$. | 11000 | Number Theory | olympiads | null | null | \(\boxed{11000}\) | Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number written with the same digits in reverse order is equal to $N$. It turns out that the number $N$ is divisible by 100. Find $N$.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{d c b a}=1000 d+100 c+10 b+a$, where $a$, $b, c, d$ are digits and $a \neq 0$.
According to the condition, $X+Y$ is divisible by 100, i.e., $1001(a+d)+110(b+c) \vdots 100$.
We have $1001(a+d) \vdots 10$, i.e., $a+d \vdots 10$, from |
Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90. | Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.
According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.
We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so since $d \neq 0$ and $d$ is a digit, $d=5$. Next, $110(b+c)+10 \vdots 100$, i.e., $b+c+1 \vdots 10$, from which, since $b$ and $c$ are digits, $1 \leq b+c+1 \leq 19, b+c=9$. Finally, $2000 a+110 \cdot 9+10 \vdots 9$, i.e., $2 a+1 \vdots 9$, from which, since $a$ is a digit, $a=4$.
Thus, $X=4000+90 b+90+5=90 q+45$. | 45 | Number Theory | olympiads | null | null | \(\boxed{45}\) | Problem 3. A four-digit number $X$ is not divisible by 10. The sum of the number $X$ and the number obtained from $X$ by swapping its second and third digits is divisible by 900. Find the remainder when the number $X$ is divided by 90.
The following text is the beginning part of the answer, which you can refer to for solving the problem:
Let $X=\overline{a b c d}=1000 a+100 b+10 c+d, Y=\overline{a c b d}=1000 a+100 c+10 b+d$, where $a$, $b, c, d$ are digits and $a \neq 0, d \neq 0$.
According to the condition, $X+Y$ is divisible by 900, i.e., $2000 a+110(b+c)+2 d \vdots 900$.
We have, $2 d \vdots 10$, i.e., $d \vdots 5$, so since $d \n |
Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number? | We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$ | 1 | Number Theory | olympiads | null | null | \(\boxed{1}\) | Problem 3. For what least natural $k$ is the expression $2017 \cdot 2018 \cdot 2019 \cdot 2020+k$ a square of a natural number?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will prove that $k=1$ already works. Let $n=2018$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1) |
Problem 3. For what least natural $k$ is the expression $2019 \cdot 2020 \cdot 2021 \cdot 2022 + k$ a square of a natural number? | We will prove that $k=1$ already works. Let $n=2020$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$ | 1 | Number Theory | olympiads | null | null | \(\boxed{1}\) | Problem 3. For what least natural $k$ is the expression $2019 \cdot 2020 \cdot 2021 \cdot 2022 + k$ a square of a natural number?
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will prove that $k=1$ already works. Let $n=2020$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1) |
Problem 3. For what least natural $k$ is the expression $2018 \cdot 2019 \cdot 2020 \cdot 2021+k$ a square of a natural number | We will prove that $k=1$ already works. Let $n=2019$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1)+1=\left(n^{2}+n-2\right)\left(n^{2}+n\right)+1= \\
& =\left(\left(n^{2}+n-1\right)-1\right)\left(\left(n^{2}+n-1\right)+1\right)+1=\left(n^{2}+n-1\right)^{2}
\end{aligned}
$$ | 1 | Number Theory | olympiads | null | null | \(\boxed{1}\) | Problem 3. For what least natural $k$ is the expression $2018 \cdot 2019 \cdot 2020 \cdot 2021+k$ a square of a natural number
The following text is the beginning part of the answer, which you can refer to for solving the problem:
We will prove that $k=1$ already works. Let $n=2019$, then for $k=1$ the expression from the condition equals
$$
\begin{aligned}
(n-1) n(n+1)(n+2)+1 & =(n-1)(n+2) \cdot n(n+1) |
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